Solution of HC Verma : Concept of Physics – I : Chapter 12 Simple Harmonic Motion

5/21/2012

162 Comments

Solution of HC Verma : Concept of Physics – I : Chapter 12 Simple Harmonic Motion

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162 Comments

Jaffar

10/25/2012 12:52:48 pm

I am not able to understand the free body diagram of q 15, chap= 12.

Answer

10/26/2012 11:40:44 am

Calculate part A
1. Equate the spring force and force due to the component of their weights.
2. from this you will calculate the value of x.

Calcuating part B
1. You may see in the free body diagram that P is the reaction force between the two masses.
2. when blocks will seperate, P=0.
3. from this we may calculate X by keeping P=0 as given in the solution.

Calculating part 4
1. for this we are using change in Kinetic Energy = Change in Potential Energy.
2. And through this equation we are calculating common velocity.

jaffar

10/26/2012 11:42:36 am

thanks for the solution

Sanjay

11/3/2012 02:32:15 pm

How we are calculating spring constant in question 21?

pooja link

6/11/2013 06:41:05 pm

sir plz solve wuestion number 8 , 10 nd 1

Admin

11/3/2012 03:47:11 pm

This is attempt to provide free education to children who are preparing for engineering entrance examination.
If you are having any doubt regarding solving questions of HC Verma or understanding any topic related to this, you may leave your comment.
we will give your answer within 24 hours.
Thank You

Aniruddha link

2/10/2014 01:02:43 am

Sir,would you please explain Q.23,While taking enery conservation why do we take in account -mgx.Thank you

PRATAP SINGH link

11/28/2012 06:47:04 am

how to get i solution of SHM

Admin

11/28/2012 10:30:46 am

Dear Pratap,
SHM is a very simple chapter.
Here I will tell you how to solve the problems of SHM
You know that a particle is said to be in SHM if it follows a = - ω²x
Where a= acceleration of the particle
ω = angular velocity of the particle
x = displacement from the mean position
You know that in most of the questions you are asked to find time period or angular velocity.
If any situation is given
1. First find the total force acting on the particle. Make sure that the force should be in terms of x.
2. Divide the force with mass.
3. Now apply equation a = -ω²x
4. Through this find the value of ω.
5. Now you may use the value of ω to find other values like time period etc.
This is the general method. If you are having problem in understanding any specific concept or numerical, then you may write to us.

Yadu

7/13/2016 11:15:01 am

Step 1.. total force means resultant force?

bhavya

11/30/2012 04:15:51 pm

i have a big problen in q15.........
pls explain 2nd part ........

& also in q 18 c part at eqb.posiion fow can pe be zero

Admin

11/30/2012 09:22:26 pm

For part B –
Please see free body diagram given in the solution.
Here we are considering second block. There are three forces that are acting along the inclined plane on second block.
1. P = It is the reaction force between the blocks
2. m₂gsinθ = It is the component of weight along the inclined plane
3. m₂a = m₂x₂ω² =It is pseudo force which is equal to restoring force of the spring, also called centripetal force.
Here x₂ = position at which blocks are seperating
Now Balancing these equations we get
P + m₂a = m₂gsinθ
We put P = 0 because blocks are loosing contact
m₂a = m₂gsinθ
Through this we will find x₂
Solution of Question 18 part c –
We know that total mechanical energy = Kinetic Energy + potential energy
At equilibrium position or mean position
Potential energy = 0 because
Potential energy = 1/2kx²
At equilibrium x = 0
So Potential energy = 0
So total mechanical energy = Kinetic energy
Total mechanical energy = (1/2)k(F/k)² = Kinetic energy

jay

1/28/2013 09:37:09 pm

how is (m1+m2)gsin(theta) /k the strings natural length?isn't it the strings equilibrium compression?

Admin

1/29/2013 04:50:26 pm

Dear Jay,
Here according the figure the spring is naturally under compression due to the masses. so, here springs equilibrium compression is the natural length.

bhavya

12/7/2012 09:34:31 pm

equilibrium position has to be the mean position.that is it is the point where f=kx.then x would be f/k.therefore there is extension of f/k . so how can x be 0???????????????

Admin

12/8/2012 10:44:39 am

Dear Bhavya,
x = f/k
Here x is the amplitude
where f = external force due to which SHM is occurring.
In order to check the mean position follow the simple rule -
1. First remove external forces due to which SHM was occurring and see that what is the natural extension(x) when spring is not moving or when the spring is at rest. There are two cases for this -
case 1 - when mean position x = 0
For Example - a spring lying horizontally on a table with box attached to its end.
now you may see that there will no extension because there is no internal force to elongate the spring as the spring if horizontal and mean position x = 0
case 2 - when mean position has some value -
for example take a vertical spring holding a box of weight W at its end.
Now you may see that there will be permanent extension due to weight W which is given by x = W/k.
Now you may see that automatically spring is resting at x = W/k which is its mean position.

Shreyas

12/19/2012 06:10:26 pm

Can u please explain why the force due to spring A and B in q19 is kx/root2...?

Admin

12/19/2012 10:41:17 pm

Dear Shreyas,
In question number we can see in the diagram that spring C is at 45° to the springs A and B. Now consider Spring C to be as hypotenuse to the springs A and B.
Now if spring C(hypotenuse) is displaced by x then spring A and B will be displaced by x cos 45° = x/(root 2).
Now force = Kx/root 2

Shreyas

12/19/2012 06:12:15 pm

and please also explain for q20.how do we get kx/2

Admin

12/19/2012 10:50:29 pm

This question is also same as above.
Here we can see that spring C is at 120° with spring A and B.
Now extend spring in the opposite direction between A and B.
we can see that it will at 60° with spring A and B.
Now again consider spring C to be as hypotenuse. if spring C is displaced by x, then spring A and B will be displaced by x cos 60° =
x/2 .
force = Kx/2

Shreyas

1/21/2014 10:12:22 pm

sir in the previous problem it works but in this question ......only force along spring A appears to coming true.....how can the force along spring B be kx/2.................plz help

suraj

12/24/2012 03:10:38 pm

pls add animation

ankita

1/8/2013 07:04:06 pm

in q.39 where's d system? i mean d diagram is lyk dat of a pendulum bt in d buk ders a ball nd a concave surface. so, 4m where omega nd omega1 came? i cudnt understand

Admin

1/9/2013 06:05:46 pm

Dear Ankita,
in this question a small ball is rolling on a concave surface. The ball is having two angular velocity. These are -
1. ω1 = This is the angular velocity of the ball about its own axis because in the question it is given that the ball is rolling purely without slipping.
2. ω = This is the angular velocity of mass of the ball about the center of curvature because concave surface is just the arc of circle of radius R and mass of ball is also rotating about the center of circle of radius R. Here mass is situated at the center of sphere. That's why we have taken radius to be as (R - r)
where
R = radius of curvature of concave surface
r = radius of sphere

manvi

1/30/2013 09:29:09 am

Can you please explain me questions no. 14. .

Admin

1/30/2013 08:46:46 pm

Dear Manavi,
you need to understand the free body diagram of this question.
please see the figure given in the solutions.
we can see that there are three forces acting on mass m
1. mg = weight downward
2. R = reaction that we need to calculate
3. mrw² = psedo force acting upward. we know that mass m is exhibiting SHM so, it will move with acceleration w²r in the downward direction because a = - w²r. we know that the pseudo force acts in opposite direction of the acceleration. Thats why it is acting upward.
Now applying vertical equilibrium we get equation 1 as shown in the above solutions and get the answer.
part b -
R is minimum when mrw² is maximum.
mrw² is maximum when the mass m is at farthest distance from the equilibrium i.e x.
put r = x and get the answer as shown in the solutions
part c -
The two blocks will oscillate together when R is greater than Zero means that when block M is exerting some reaction force on m. if R is zero then this means that they are about to separate.so, for the two to oscillate together R must be positive. For the maximum amplitude R must be minimum so we have taken R minimum = 0
in eq 1 and solved the answer.

manvi

1/30/2013 11:24:44 am

Please explain me how can we apply energy method and also explain me questions no. 25.

Admin

1/30/2013 08:16:39 pm

Dear Manvi,
First of alll i would like to tell you about the energy method.
Energy can be applied everywhere.
There are two situations. These are -
1. when there is no loss of energy
2. when there is loss of energy

1. when there is no loss of energy then we an say that total mechanical energy remains conserved.
Let us take an example -
Take two points as A and B. The object has moved from point a to b.
then according to the energy mothod
mechanical energy at A = mechanical energy at B = constant

2. Now when there is loss of energy
then by energy method we get
mechanical energy at A = mechanical energy at B + energy lost

Now i am giving the solution for question 25 -
we know that the masses moved by distance x opposite to each other. so, total compression = 2x
Now as there is no loss of energy because all the forces are conservative.
so, total mechanical energy = constant = c
(1/2)kx² + (1/2)mv² + (1/2)mv² = c
Now we are derivating it with respect to time, we get
a = -2kx/m
we know that
for SHM, a = - w²x = -2kx/m
now we can solve as done in the solutions

sumu

1/31/2013 03:10:21 am

Q. two paricles are under going in SHM.with same amplitude A and same time period T.
what is the phase difference between the particles when they are at A/2..????

Admin

1/31/2013 08:40:56 pm

Dear sumu,
For the phase difference we need to know the initial phase difference between the two particles and also the the angular velocities of the particles. if these are given in the question then tell me because these are to be taken in the consideration or else the question will be incomplete.

manvi

2/2/2013 12:57:25 pm

Page 246 question 12 fig. assume Both the mass m. Each end of the spring is stretch by force f.if the mass are released , what is the period of oscillation .

Admin

2/3/2013 09:15:40 am

Dear Manvi,
This question is same as that of question 25 of exercise. The only difference is that instead of moving inward they are moving outward. You solve this according to the method given in the 25 question you will get the answer. I think the answer will be same as that of 25.

manvi

2/3/2013 11:50:49 am

Thanks.

manvi

2/3/2013 12:00:17 pm

Vertical displacement of a plank with a body of mass m on it is varying according to law y=sin omega×t + root3cos omega×t. The minimum value of omega for which the mass just breaks off the plank and the moment is occurs first after t=0, are given by:(y is positively upward)

Admin

2/3/2013 08:47:23 pm

Dear Manvi,
We know that whenever a particle is exhibiting SHM, then acceleration a = - w²y
restoring force acting on particle = -mw²y
let R be the reaction force due to the plank on the body.
so, mass will only break off when the plank is moving downward because in this case the reaction force will be acting upwards.
So, we get
R = mg - myw²
At break off , R = 0
mg = myw²
w² = g/y
so, w = root(g/y)
so, w will be maximum when y is minimum
y = sin w×t + root3cos w×t
We know that wt = θ
y = sin θ + √3 cos θ
for y to be minimum,
dy/dθ = 0
dy/dθ = cos θ - √3 sin θ = 0
tan θ = 1/√3
θ = 30˚ or 150˚ or 330˚
At θ = 150˚ and 330˚ we are getting magnitude of y as minimum
magnitude of y minimum = 1
w = √10
t = 150˚/√10 or 150π/(180√10) sec

manvi

2/3/2013 12:05:12 pm

A particle perform shm in a straight line . In the first sec. Starting from rest, it travel a distance a and in the next sec. It travels a distance b in the same direction. The amplitude of shm is:

Admin

2/3/2013 09:54:06 pm

Dear manvi,
I am still solving the question. I will tell you as soon as possible.

manvi

2/3/2013 11:50:42 pm

Its answer is 2×asquare by 3×a _b.

manvi

2/3/2013 01:15:16 pm

What will be the time when 2 particle cross each other if amplitude and time period for both the particle are equal and at t=0 one particle is at +a and the other at x= _a by 2 and also they are coming toward each other.

Admin

2/3/2013 09:16:29 pm

Dear Manvi,
Please visit the link "http://physicsgoeasy.blogspot.in/2012/02/circle-of-reference-in-simple-harmonic.html". I am explaining this question with respect to the figure given on this page.
See according to the condition there are two particles. These are
1. first particle - It is making 90˚ with respect to the horizontal.
2. second particle - It is making - 30˚ angle with respect to the horizontal.
Now both are coming towards each other. This means that first particle is moving in clockwise direction and second particle is moving in the anticlockwise direction. Now total angle between them = 90 + 30 = 120˚.
Let time taken by them to cross each other = t
angular velocity of both particles = w
angle covered by them in time t = wt
2 wt = 120˚
wt = 60˚
t = 60˚/w

manvi

2/4/2013 02:09:03 am

I don't know how to open the link

Admin

2/4/2013 08:55:59 pm

Dear Manvi,
Please copy this link and paste it in the address bar (where you type the name of website) and enter it.

ayushi link

2/3/2013 11:10:29 pm

i am facing a lot of problem in energy conservation in shm

Admin

2/4/2013 09:44:07 pm

Dear Ayushi,
you just start solving questions related to that and tell me in which question you are facing problem. I will explain you according to that context.

sreerag kv

2/16/2013 11:55:52 am

sir,in question,54 in shm at the point of finding w^2(w square)..
how the 2 came at the denominator..??? when i did halves from both side get cancelled and while applying value of m.inertia i got 2 at numerator.....?? plz answer me...

Admin

2/17/2013 08:03:23 pm

Dear Sreerag,
Yes you are right that 2 must come in the numerator.
There is one more mistake in the solution that While finding T2 we must use the radius as L/2 instead of L.
T2 = √ ( (mrw²)² + (mg)² )
Now use r = L/2 in the above formula.

Rajeev verma

2/22/2013 07:19:45 pm

Sir please explain me question 31 briefly

saima

3/7/2013 03:22:01 pm

if there is a planet whose daimeter and mass is half of earth's..
the day temperature is 800K. is oxygen gas possible at this planet????
(R= 1.38*10^-23J/K.) ( mass of one molecule of oxygen= 5.3*10^-26 kg)

theyatin

3/18/2013 07:32:30 pm

dear saima,
this question is irrelevant to this chapter you need to use gravitational force and thus escape velocity of the planet and then you will have to calculate kinetic energy of the gas oxygen
Escape velocity, ve = √2GM/R

Let vp = escape velocity on the planet.

ve = escape velocity on the Earth.

vp/ve =√(MP/RP × RE/ME) = √(1/2×2/1) = 1

vp = ve = 11.2 km/s.
From kinetic theory of gases

vrms = √(3RT/M) = √(3NKT/M) = √(3NKT/Nm)

where N = Avogadro's number

m = mass of oxygen molecule

K = Boltzmann constant

vrms = √3RT/m
here m is mass of molecule of oxygen please calculate rest of these by yourself

theyatin

3/18/2013 07:37:53 pm

in case Vrms is smaller than that of Vescape then molecule can not escape the planet

saima

3/7/2013 03:23:51 pm

sir please reply soon.. thank you..

Tejasv

3/17/2013 02:38:30 am

Sir, I am finding a great problem in finding amplitude . Is there any method to find it easily .

XZCVB

4/9/2013 10:54:35 pm

u r grr8

theyatin

4/10/2013 01:26:54 am

dear,
rather than being sarcastic just focus on study . . .

alisha

3/12/2014 04:26:02 pm

Well sir.. I don't think @xzcvb was trying to be sarcastic.. he/ she was just thanking u for taking the time to help out students wid hcv..

sonam link

3/21/2013 01:43:21 am

i am facing problem in this ques.? the bob of simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t in air. neglecting frictional force of water and given that the density of the bob is (4/3)*1000kg/m3 . what relationship t and t is come?

sanjana

4/9/2013 03:31:37 pm

In the question no 18 when we a constant force is applied, then can we compare this shm to the shm of vertical spring mass system??

if we can then according to conservation of energy we get
at the extreme positions
1/2kx<power 2> = mgx
x =2mg/k
or in this case
x=2F/k
the ans is F/k

theyatin

4/9/2013 05:49:03 pm

dear sanjana,
in case of two masses combined with a spring this equation may justify. . .
but here only one mass is under consideration
and F=kx is constant of elasticity into displacement.

Nitin

4/10/2013 01:32:34 am

Dear theyatin,
this equation may hold good then as well,
But considering just a spring-mass system in a vertical plane then the as well the max amplitude is = 2mg/k. derived from the above equation .
I think what sanjana means is that, mg is just a constant force, which is just like the F force,
The only diff that may have occurred was when the surface was rough, but as we know it is nt.
Just thinking out loud, say a block m is hanging, from its natural length in the vertical plane, and another body of the same mass , is being pulled by a constant force F.
There seem to be no diff between the two cases, besides the axis itself

brar

5/3/2013 02:58:08 am

not clear with question 20 sir can u please explain more elaborately

theyatin

5/3/2013 03:05:06 pm

dear,
20th Q is same as of 19th only diffeence is that each spring is 120' from each other so the motion would be more symmetric.
now mass is pushed against c so the total force would be due to
C and resultant of A & B
F=Fc + Fa,b
F=m.a
thus A=F/m putting value of F from above
angular velocity w^2 = a/x
and T=2*Pi/w

Rakesh

6/7/2013 02:28:22 pm

sir in the formula for linear SHM why omega is used as omega stands for angular displacement n this is linear SHM

theyatin

6/7/2013 05:35:00 pm

dear,
what are you actually trying to ask??

rakesh

6/7/2013 05:57:22 pm

I just want to ask that if the moment made is linear then where does it make angle.

theyatin

6/8/2013 02:16:40 am

dear,
as the motion is along only one dimension thus we can say it as linear SHM but actually the motion is curved. . .
so it makes angle with every previous position w.r.t. time.

rakesh

6/8/2013 02:40:11 am

thanks sir

Aymanzoor

10/30/2013 08:42:39 pm

Sir in q14 part b
if N=mg-(mkx/M+m) dosent that mean mg and Kx are in opposite directions but how is this possible
If the block is above the eqb position that would cause the spring to pull the block downwards in the direction of mg and thus N = mg+ (mkx/M+m)
AM i Wrong

theyatin

10/31/2013 02:22:41 am

dear,
the spring would oppose the force due to weight by the force caused by spring constant.

Aymanzoor

10/30/2013 08:49:43 pm

Sir in q26
why is theta= x/L
and not x/[L+(a/2)]
as the distance b/w the com and the assumed point of suspension is L+a/2

Aymanzoor

11/2/2013 04:31:23 am

Sir please reply soon

subanshu

11/19/2013 02:00:23 am

Question 2 why talking value of pie 3.14

theyatin

11/19/2013 04:51:55 pm

dear,
we can use 22/7 or 3.14 as per our convenience but you need to specify as your answer may vary a little.

subanshu

11/19/2013 09:15:55 pm

actually sometimes the value is taken 180 radians and sometimes 3.14 when and which cases the above values are to be substituted
plz do elaborate

theyatin

11/22/2013 02:17:29 am

dear,
in case you need to consider angles you take pi as 180 radians
but in case you need it in calculations like area or volume you take it as 3.14

subanshu

11/22/2013 04:33:54 pm

But only in shm finding omega valu of pie is being considered as 180 and finiding time period is being taken 3.14

theyatin

11/24/2013 01:51:06 pm

dear,
see in case you choose value of pie as 180radians for time period
then units of time period would be radians-sec
so we must consider pie as constant

shubhu

11/29/2013 12:57:58 pm

In Q. No 15 part b can we say that the separation distance is independent of initial compresion of the spring

theyatin

12/2/2013 02:30:17 am

dear,
yes sort of.

Nithin

12/5/2013 11:20:09 pm

i cant understand quuestion no 56... please make it clear.....

shubhu

12/6/2013 06:01:18 pm

In Q 40 amPlitude comeS out Be √2R but Since ther is No field outside the surface them how it can go to max. Displacement of √2R ??

shubhu

12/6/2013 06:04:52 pm

In Q 41 amPlitude comeS out Be √2R but Since ther is No field outside the surface them how it can go to max. Displacement of √2R ??

shubhu

12/7/2013 12:10:22 pm

In Q 51,52,54 when applying energy conservation why we are not considering kinetic energy of the system ???? Why only rotational K.E???

shruti

12/20/2013 11:36:55 pm

pls explain q.26. first step...
and q16 part (f)

shubhu

12/21/2013 02:15:47 am

Dear for 26 just treat it as a simple pendulum analysing its centre of mass only! Don't get confused with the dimensions given!!!
For 16-f- just apply energy conservation considering left extreme and right extreme as end points !!

shubhu

12/21/2013 02:17:17 am

shubhu

12/21/2013 02:17:29 am

Ansh Arora

12/23/2013 01:51:05 pm

In Q 25. if we take total extension of spring 'x' ie - displacement of each block is x/2. Then the total energy of the system is 1/2k(x^2) + m(v^2).
Differentiating this, i am not able to get the answer .

himanshu

12/25/2013 04:32:16 pm

Sir in SOLVED EXAMPLE no 10, plz explain why time period of the spring is taken half the real value?, oscillation is complete, then why take half of the time period?

theyatin

12/27/2013 03:38:59 am

dear, because upto the natural length of spring the block has acquired its original velocity
so we need to take half of its amplitude of its harmonic motion. that is half a period

John Doe link

12/27/2013 08:52:10 pm

In question no. 20 (all at 120 degrees) can we apply the concept of springs in series and parallel.. Using that im getting keffective as 2/3 k and the correct answer but im not really sure if its the correct way.
Also i can apply this to cases with different angles like 19.

Thank you!

theyatin

12/30/2013 12:27:12 am

dear,
in this case you can but in case angle is acute or there are many strings this would be tougher so its your choice to choose way to solve.

sai

1/2/2014 02:47:18 pm

sir in ques 1 2nd part ...u used equation a=omega square*x
but d acctual equation is a = minus(omega square0 * x..
where did d minus sign go

theyatin

1/3/2014 12:49:15 am

dear,
its given
a= - w^2 x
look again

parima

1/7/2014 10:24:49 am

In q16 why are b and e part different ?

theyatin

1/8/2014 02:13:12 am

dear,
because in part e)
we need to find energy at left extreme
so there is no kinetic energy thus only potential energy and that is equal to the force applied.
that is P.E.=F=1/2*k *x^2

sumit

1/13/2014 01:31:22 am

sir ques 16 part b

theyatin

1/14/2014 01:54:55 pm

dear
here in this Q
total mechanical energy =p.e. +k.e.
here p.e,=force applied on the string=1/2*k*x^2
and k.e. is given as it is

vish

1/25/2014 12:17:42 pm

sir in objective 1 ques. no 21 is confusing me if i am also observing from same frame than would not the oscillation be same?

Abhishek

1/26/2014 06:22:46 pm

What's the main concept behind angular shm

Ankit

2/10/2014 03:16:01 am

Sir,
how can i know the phase angle if the graph is given?

theyatin link

2/19/2014 01:21:41 am

dear,
this link may help

tnv

2/18/2014 06:35:50 pm

can you please explain question 31 ? im confused in R1 + R2 =mg part!!

tnv

2/18/2014 06:47:14 pm

never mind i solved the problem just now

tnv

2/20/2014 03:48:11 pm

sir! i am not able to understand the last 2 steps of the 54th sum .

Ansh Arora

2/21/2014 07:48:00 pm

Can you please solve q no.39 by force method.

anisha

2/28/2014 08:57:40 pm

Sir,plz explain q18)b)

Admin

3/6/2014 04:00:57 pm

Please wait. Yatin sir will resume work within 2-3 days.

alisha

3/7/2014 01:10:14 pm

Sir in ques 19 why did we not consider mg force (the weight of the body).. we solved only wid the sping forces.??

mayank

3/14/2014 05:39:25 pm

can you provide more advanced shm questions

theyatin

3/17/2014 02:22:41 pm

dear, right now we are not having additional problems we will upgrade levels at time
you can ask your problems here till then

sachin

3/20/2014 11:16:41 pm

I can't understand the solution of que no 31. Plz help me...

Rohan

4/9/2014 03:12:30 pm

In question no:14 why did we replace w^2 by (k/M+m) ?

theyatin

4/12/2014 11:29:00 pm

What marks are you expecting in iit mains

swadhin agrawal link

4/22/2014 05:23:27 pm

Sir can you plz explain me the concept behind
Composition of two SHM,damped harmonic motion & forced oscillation & resonance.

syed sohail sultan

5/14/2014 04:00:32 am

suppose there is two identical sphere of radius r,one is filled with an ideal liquid and other is a solid body.will the time period of the two physical pendulam be same?sir plz explain the answer.....

shubhu

5/14/2014 02:47:26 pm

dear both the cases are of compound pendulum so we have to use T=2pi√I/mlg
In first where we have ideal liquid filled so its non viscous fluid which can't rotate so it has only transltational motion ! So its I =ml² [ m be mass of water and l be the distance of hinge from centre of sphere ] in second its solid body so I=2/5mr²+ ml² !!! So its clear they are not equal!!

Trishnika Mukherjee

5/16/2014 05:53:40 pm

Sir will you kindly explain Q15 part c elaborately.?

ER.MIRJAIK

5/20/2014 01:47:39 am

ER.MIRJAIK

5/20/2014 01:49:49 am

5/22/2014 01:08:38 am

Sir plz explain objective 2 q no 15 option A

mirjaik

5/31/2014 12:12:52 am

In Q. No 15 part b can we say that the separation distance is independent of initial compresion of the spring

niti n

5/30/2014 06:11:00 pm

kindly explain Q.9 of objective ...mathematically...plz...

mirjaik

5/31/2014 12:21:47 am

dear
here in this Q
total mechanical energy =p.e. +k.e.
here p.e,=force applied on the string=1/2*k*x^2
and k.e. is given as it is

admin

6/22/2014 01:45:26 pm

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Dmgcfhfrvch link

10/3/2014 04:59:00 pm

Diff noggin,

sanjay

11/15/2014 09:14:06 pm

Please explain direction of forces in no. 15

Prachi Singhal

11/24/2014 02:14:38 am

Please explain Q. 22.
How we take direct kx1=mg, kx2=mg, kx3=mg
Why we dont take y1+y2+y3=y
where y1,y2,y3 are extension made by spring of spring constant k1,k2,k3 respectively.

Thank you..

JATAN

11/30/2014 01:16:13 pm

Sir Please explain me the 1st question of objective 2 chapter-12 simple harmonic motion hc verma part 1. I think that "c" option is also correct.

ping

12/15/2014 01:59:24 pm

A scale of a spring balance that reads 0 to 15 kg is 12 cm long. A package suspended from the balance is found to oscillate vertically with a frequency of 2 hz. What is the spring constant and mass of the package

ArnabBiswas

12/29/2014 04:39:26 pm

sir q16d ande ihavnotunderstood

vikash yadav

1/2/2015 12:58:55 am

sir please tell me how can we solve proble
m related time period

Jaikirat Singh

1/12/2015 11:57:54 am

Sir, Can you plz explain me how can I solve the third part of Q16 by Calculus.

tanu gupta

1/23/2015 03:54:13 pm

sir can u plz explain q37

sattar

1/25/2015 12:20:39 am

How to solve problems easily on Rotational Dynamics...i didn't understand it properly...is it that we need to solve more no of problems to get clarity or conceptual understanding is imp before solving problems...plzz help me out

Gnanaram

1/25/2015 07:51:46 pm

I can't understand question 12 in SHM CHAPTER 12

anirudh bagla

2/13/2015 12:36:31 am

q 27 why cant conservativeon of energy be applied

Hemavathi

3/1/2015 08:57:45 am

will u pls explain me the 2nd part of 1st qs..??

suyash

3/3/2015 04:48:51 pm

i did not uderstood calculation in question 54

shubhi

4/15/2015 02:12:30 am

Sir plz explain ques no. 39.

4/17/2015 07:51:15 pm

Q 57 we have to do vector so why

prerna

5/20/2015 02:51:05 pm

How to solve no.37 in shm hcv

Madhavi

10/1/2015 10:06:39 pm

Can you please tell me how to solve question no. 24 by force method??

sameer maurya

10/31/2015 04:17:46 am

can you tell me Q.no 14

Hillol das

11/11/2015 04:09:14 am

Sir can u plz explain q 10 & 11 of objective 1 part of chap 10

anupam

1/17/2016 09:50:30 pm

qUes.14

anupam

1/17/2016 09:50:52 pm

qUes.14

Rohan

1/31/2016 05:25:24 am

Pls explain q15 part c in detail (the potential energy part)

J Mukherjee

2/1/2016 10:25:41 pm

I would like you to help me solve the following question .
A wooden block of mass .01 kg executes a SHM with amplitude of .02 m under the action of two identical springs of spring constant of magnitude 0 .05 N/m . The springs are tied at two opposite

prateek

2/4/2016 08:19:00 am

Shambhav laddha

3/1/2016 04:31:17 am

Shambhav LADDHA

Sir plz tell that how to calculate the 'I' for given system like in Q-48

Shambhav

3/2/2016 05:05:02 am

Plz reply to above oue

Jayshree link

3/4/2016 06:57:50 am

How can solve q 15

Taral jain

12/12/2016 09:31:47 pm

Please solve question number 37

Surya

12/17/2016 11:08:06 am

In shm of hcv , in ques. 31 , are the direction of the wheels wronglly marked

Anurag Atul

12/28/2016 08:52:51 am

Actually I am not able visualize how it will oscillate and how COM concept is used. Please give the figure also.

Jai

1/27/2017 12:51:59 am

vertical displacement of a plank with a body of mass m on it is varying according to the law y=sinwt+3^1/2coswt. the minimum value of w for which the just break off the plank and the moment it occurs first after t=0 are given by :

Abhishek

2/15/2017 08:58:57 am

Explain q.30 with solition .........plz .

Dhanshree

3/28/2017 11:12:37 pm

I don't understand question 20. An elaborate solution is needed . Please help!

Sukanya

4/3/2017 08:07:40 am

Sir, please, how is the equation in ques 54 solving out to be kθ²/2I?
Iω²/2=kθ²/2
=> ω²=kθ²/I
How is the factor of 1/2 coming?

Nilam

7/25/2017 07:05:40 pm

In exercises qustion no. 30. In the solutions there is an X used. What is this x?

shreya

12/27/2017 10:25:43 pm

Please explain question no-54 of SHM.

aditya

12/29/2017 05:32:32 pm

assume that a tunnel is dug along a chord of earth,at a perpendiular distance R/2 from the earth 's centre where R is the radius of the earth the wall of tunnel is frictionless . find the resultant force on the particle

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Solution of HC Verma : Concept of Physics – I : Chapter 12 Simple Harmonic Motion

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