# Time period of SHM

vissh
Time period of SHM !!

ellloO ^.^

## Homework Statement

<Q>A particle executes SHM under the restoring force provided by a spring. The time period is T . If the spring is divided in 2 equal parts and one part is used to continue the SHM, The time period will
<A> remain T <b> become T/2 <c> become 2T <d> become T/(21/2)

## Homework Equations

>If on a particle of mass "m" at a displacement "x" from the mean position , a force " -kx " acts ;then the particle will be in SHM abt that mean position. [ k = constant]
T = [2 $$\pi$$ (m)1/2]/(k)1/2

## The Attempt at a Solution

>>Earlier, the particle was acted upon by force of magnitude "kx" when it was "x" away from mean position and thus time period
T =[2 $$\pi$$(m)1/2]/(k)/2
>>If we use the half part only , still the particle will be acted upon by force of magnitude "kx" and time period T will not change.
>>But the answer is <d> become T/(21/2)
Little help ^.^
{ lol don't know why the Pie symbol is flying above all other words }

Mentor

Consider a spring with spring constant k. You paint a small mark on the spring at the halfway point so that you can follow the motion of the center. Now the spring is compressed by an amount x. How far did the mark move? What is the compressive force all along the spring (if you were to insert a little force meter into the spring at any given point along its length)?

vissh

>Hmmm Don't know where the center of spring will be ;)
>But I know the Tension in the spring will be "kx" if the extension or compression is "x" .
You refering Tension as the compressive force or u talking abt smthing else??
>But still I do not got why the time periods will not be same Whether whole spring is taken or half part (of it) is taken as ""the time period"" for this spring-mass system depend on spring constant "k" and not on spring's length {As much as i know}
>>>Any ideas anyone >:D (^_^)

Mentor

Springs compress uniformly. If the end of the spring moves by x, how far does the center move?

vissh

Hmmm uniformly... If i keep this in mind, the center of spring will move by a distance "x" {But then how the "spring" gets compressed if every point moves the same distance xD. I think I got "uniformly" wrong way .But can't think of what else would happen :( hehe }
Can u help me out more ^.^

Mentor

'Springs compress uniformly' means that all of the spring participates evenly in the compression -- they don't squeeze tightly at one point and remain loose at others.

Draw a picture of a spring standing on end. Beside it draw a picture of the same spring compressed by an amount x. Draw a line connecting the tops of the springs. Draw a line connecting the centers of the springs. How much did the center move compared to the top?

vissh

Hmm by my rough diagram, I got the spring's center will move less than spring's top.
If the top moved by "x" , the center moved by "x/2" {approx}

Mentor

Okay. In reality the center will move exactly by half the distance.

Now, the other thing to note is that the force applied to the spring is felt all along the length of the spring, and is finally transmitted to whatever is holding the fixed end. Anywhere along the length the compressive force is the same. If you cut the spring at the halfway point and measured the force there, it would be the same as the applied force at the top end.

Now, given those two facts, what's the 'k' for the half spring? The force is the same as before, but the distance is halved...

vissh

Oo oO :O hehe
>Got the spring constant for half part to be "2k" and thus the time period will be
"T/(2)1/2".
>I thought that spring constant depends only on the material from which the spring is made and not on its length. Thanks to you got a new thing to know :D