SHM Help

I need help for the second part please. How do the oscillations being damped prevent them from being S.H.M. as well the other 2 points? I need an explanation so I can understand. The oscillations would still have the same time period eve if they are damped.

Q+MS 2.png

Low frequency due to mass/density (of spheres)

Can someone please explain that to me as I don't understand it.

Q+MS 3.png

Isn't T supposed to be constant as it is independent of the amplitude theta, right?

T=2pisqrt(m/k)
So surely as m increases T increases like the graph of y=sqrt(x)?

How does good suspension in a car help prevent resonance in the various parts of the car?
Prevention of resonance:
Damps oscillations (1)
Fewer forced oscillations (1)
Explanation of damping [e.g. in terms of energy transfers] (1) Max 2

For the last part, I understand that the suspension damps oscillations, but I'm not fully sure how there are fewer forced oscillations. Is it because that the damping causes the oscillations to die away quicker, so they stop quicker. Hence there are fewer forced oscillations?

Q+MS 5.png

I think the MS answer is wrong as I got k=1.40Nm^-1 using T=2pisqrt(m/k). What did others get?

THANKS SO MUCH! :biggrin:

(edited 10 years ago)

Reply 1

spiruel

Skipping the resonance questions as I don't think I'm able to explain them:

1 - I think it can't be classed as SHM because energy is being lost, and therefore the oscillations cannot continue indefinitely.
2 -

1 = 2\pi\sqrt{\dfrac{l}{g}}

only applies when there is a small angle. Where

\lim_{\theta \to 0}sin(\theta) = \theta

applies. This is because the restoring force requires a straight line motion of a pendulum for the formula to work. See:

F = -kx

Using the the correct formula, you see mass isn't a factor and therefore has no affect on T.
3 - It's a pendulum so you can't use the mass-spring formula

T = 2\pi\sqrt{\dfrac{m}{k}}

.
4. I got

Unparseable latex formula:

1.4Nm$^{-1}$

as well!

(edited 10 years ago)

Reply 2

Stonebridge

13

1. Damping introduces a second force. Friction. The definition of SHM is very strict. The restoring force on the object must be proportional to its displacement from the equilibrium position. Any other forces acting mean it's not SHM. Friction opposes the restoring force. It's a matter of definition.
If you damp an oscillation you increase its period. Always. If the damping is small the change in T isn't very much. This is why you get away with calling a mass on a spring in air SHM, because the damping is very light. (Try and get it to oscillate in oil!) Strictly speaking, it's only SHM if the motion is frictionless. (And the spring must obey Hooke's Law.) Just use your common sense. If you increase the damping of an oscillation, you will eventually prevent the oscillation altogether and the period will be infinitely large and it will never complete a cycle. (This is called heavy damping.)
I think you are assuming that anything that oscillates is SHM. This isn't the case. SHM is a very special case of oscillation. However, if you know something is SHM, or very nearly SHM, there's lots of lovely maths that can help you describe the motion and calculate useful things like period and frequency.

2.
The rubber barrier will oscillate as it's an elastic material. It has its own mass so the frequency at which it oscillates will be determined by that mass and the spring constant of the material. If you add lead balls you increase the mass of the barrier and therefore reduce its frequency of oscillation. The formula for the frequency of oscillation of a mass subjected to an elastic force says that if you increase the mass you reduce f and increase T.
If the barrier vibrates at a lower frequency it will resonate at a lower frequency. If this is the case it will resonate for lower frequency sound. (The low bass notes.) This means these low frequency bass sounds will tend to be absorbed by the barrier and not pass through it.

Reply 3

krisshP

OP

14

Original post by Stonebridge

X

WOW, it makes some nice sense now when you put it like that.

THANK YOU! :biggrin:

PRSOM

Any idea for the other question in the OP? This one:

How does good suspension in a car help prevent resonance in the various parts of the car?
Prevention of resonance:
Damps oscillations (1)
Fewer forced oscillations (1)
Explanation of damping [e.g. in terms of energy transfers] (1) Max 2

For the last part, I understand that the suspension damps oscillations, but I'm not fully sure how there are fewer forced oscillations. Is it because that the damping causes the oscillations to die away quicker, so they stop quicker. Hence there are fewer forced oscillations?

(edited 10 years ago)

Reply 4

krisshP

OP

14

Original post by spiruel

3 - It's a pendulum so you can't use the mass-spring formula

T = 2\pi\sqrt{\dfrac{m}{k}}

.

http://www.thestudentroom.co.uk/attachment.php?attachmentid=259817&d=1388506557

I read up a bit on SHM. Yeah,

T = 2\pi\sqrt{\dfrac{l}{g}}

applies for a simple pendulum with small amplitude to allow for the assumption its proof that sintheta=theta where theta is in radians. However

T = 2\pi\sqrt{\dfrac{m}{k}}

applies for a mass on a spring. Hence I must use

T = 2\pi\sqrt{\dfrac{l}{g}}

in this case. for the LHS graph. But I still don't understand why the graph is not completely straight considering that T is independent of amplitude theta from the formula :confused:

. I do understand the RHS graph though as in

T = 2\pi\sqrt{\dfrac{l}{g}}

T is independent of m.

Thanks a lot helping.

Reply 5

spiruel

Original post by krisshP

http://www.thestudentroom.co.uk/attachment.php?attachmentid=259817&d=1388506557

I read up a bit on SHM. Yeah,

T = 2\pi\sqrt{\dfrac{l}{g}}

applies for a simple pendulum with small amplitude to allow for the assumption its proof that sintheta=theta where theta is in radians. However

T = 2\pi\sqrt{\dfrac{m}{k}}

applies for a mass on a spring. Hence I must use

T = 2\pi\sqrt{\dfrac{l}{g}}

in this case. for the LHS graph. But I still don't understand why the graph is not completely straight considering that T is independent of amplitude theta from the formula :confused:

. I do understand the RHS graph though as in

T = 2\pi\sqrt{\dfrac{l}{g}}

T is independent of m.

Thanks a lot helping.

T is independent of theta assuming small angle approximation, as I mentioned earlier. As theta increases this is not the case, and T begins to increase. Hope that makes sense!

Reply 6

krisshP

OP

14

Original post by spiruel

T is independent of theta assuming small angle approximation, as I mentioned earlier. As theta increases this is not the case, and T begins to increase. Hope that makes sense!

Why an increase in T?

Reply 7

Stonebridge

13

Original post by krisshP

Why an increase in T?

If you want the exact answer you have to do the maths for the case where the restoring force is no longer directly proportional to the displacement but varies with the sine of the angle. It isn't SHM any more.
If you want a hand waving intuitive answer, it's because by increasing the amplitude you are increasing the total distance the pendulum has to swing. However, because of the fact that the restoring force varies with sine (not directly prop and not SHM as discussed before) it doesn't increase fast enough to compensate for the extra distance travelled and doesn't give the mass enough acceleration and speed to keep the period the same. So the period increases.
As I say. It's in the maths. Sometimes in physics if you want to know "why" you just have to plough through the equations.

(edited 10 years ago)

Reply 8

krisshP

OP

14

Original post by Stonebridge

If you want the exact answer you have to do the maths for the case where the restoring force is no longer directly proportional to the displacement but varies with the sine of the angle. It isn't SHM any more.
If you want a hand waving intuitive answer, it's because by increasing the amplitude you are increasing the total distance the pendulum has to swing. However, because of the fact that the restoring force varies with sine (not directly prop and not SHM as discussed before) it doesn't increase fast enough to compensate for the extra distance travelled and doesn't give the mass enough acceleration and speed to keep the period the same. So the period increases.
As I say. It's in the maths. Sometimes in physics if you want to know "why" you just have to plough through the equations.

Ah okay, make sense.

Thanks a lot for all the help! :smile:

Reply 9

06awaism

1

Hi guys, rather than posting another thread, I have decided to post my question on this thread as it is relevant and may be beneficial for some. Apologies for any inconvenience caused.

Anyways, I am stuck at this question. I have the mark scheme but it is a little difficult to go through. Could someone explain to me how you would go about doing this question, and mention few facts regarding the length of the pendulum as how it may cause a difference to the system if the length of the pendulum is shortened/elongated. Also, will the time period be different for both of these different pendulums due to their length? Thank You.

ImageUploadedByStudent Room1388590460.583047.jpg

ImageUploadedByStudent Room1388590460.583047.jpg

(edited 10 years ago)

Reply 10

krisshP

OP

14

Original post by 06awaism

Hi guys, rather than posting another thread, I have decided to post my question on this thread as it is relevant and may be beneficial for some. Apologies for any inconvenience caused.

Anyways, I am stuck at this question. I have the mark scheme but it is a little difficult to go through. Could someone explain to me how you would go about doing this question, and mention few facts regarding the length of the pendulum as how it may cause a difference to the system if the length of the pendulum is shortened/elongated. Also, will the time period be different for both of these different pendulums due to their length? Thank You.

ImageUploadedByStudent Room1388590460.583047.jpg

T=2π√(l/g) for a pendulum

Time for shorter pendulum=time for longer pendulum
5T_s=3T_l

Now just sub in T=2π√(l/g)

Reply 11

tmorrall

Helpful to think of the periods of both pendulums as a ratio i.e 3/5 of the period of the longer pendulum is equal to one period of the shorter since it completed 5/3 times as many cycles as the longer. ImageUploadedByStudent Room1388596555.627791.jpg

ImageUploadedByStudent Room1388596555.627791.jpg

Posted from TSR Mobile

SHM Help

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you