Time Period of SHM: T/(2\sqrt{1/2})

In summary, the time period of a particle executing SHM under the restoring force of a spring is determined by the spring constant and mass. If the spring is divided in half and only one half is used to continue the SHM, the time period will become T/(2)1/2. This is because the center of the spring will only move half the distance as the top, and the spring constant for the half spring is twice that of the full spring. Additionally, the spring constant depends on the material of the spring, not just its length.
  • #1
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 [tex]\pi[/tex] (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 [tex]\pi[/tex](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 }
Physics news on Phys.org
  • #2

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)?
  • #3

Thanks for replying :D
>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 referring 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 (^_^)
  • #4

Springs compress uniformly. If the end of the spring moves by x, how far does the center move?
  • #5

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 ^.^
  • #6

'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?
  • #7

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}
  • #8

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...
  • #9

Oo oO :O hehe
>Got the spring constant for half part to be "2k" and thus the time period will be
>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

1. What is the meaning of "T/(2\sqrt{1/2})" in the context of SHM?

The equation T/(2\sqrt{1/2}) represents the time period of simple harmonic motion (SHM). It is the time taken for a particle to complete one full oscillation or cycle in its motion. The value of T/(2\sqrt{1/2}) is dependent on the amplitude and frequency of the oscillation.

2. How is "T/(2\sqrt{1/2})" related to the frequency of SHM?

The frequency of SHM is the number of cycles or oscillations completed in one second. It is inversely proportional to the time period, meaning that as the frequency increases, the time period decreases. This relationship is represented by the equation f = 1/T, where f is the frequency and T is the time period.

3. Can the value of "T/(2\sqrt{1/2})" change during SHM?

Yes, the value of T/(2\sqrt{1/2}) can change during SHM. It is dependent on the amplitude and frequency of the oscillation. If these factors change, the time period will also change. For example, if the amplitude increases, the time period will also increase.

4. How is "T/(2\sqrt{1/2})" calculated?

The value of T/(2\sqrt{1/2}) can be calculated using the equation T = 2π√(m/k), where T is the time period, m is the mass of the oscillating particle, and k is the spring constant of the system. This equation is derived from the mathematical expression for SHM, x(t) = A cos(ωt), where A is the amplitude and ω is the angular frequency.

5. What is the unit of measurement for "T/(2\sqrt{1/2})"?

The unit of measurement for T/(2\sqrt{1/2}) is seconds (s). This is because the time period represents the time taken for one full oscillation to occur, which is measured in seconds. It is important to note that the unit of measurement may vary depending on the units used for mass and spring constant in the calculation of T/(2\sqrt{1/2}).

Suggested for: Time Period of SHM: T/(2\sqrt{1/2})