# Thread:

1. ## simple harmonic motion on a string

Hi
I need some clarification on this question
A light elastic string of natural length 50 cm and modulus 20 N has one end fastened to O on a smooth horizontal table. The other end has a particle of mass 0.4 kg attached to it and the particle is released from rest at a distance 60 cm from O. Find the time it takes to reach O.
While the string is stretched
T=kx/l
=20x/0.5
=40x
40x=-.4a
So,a = -100x
w=10
T=2pi/10 =pi/5
I get stuck from here. Can anyone help please? ThanksHi
I need some clarification on this question
A light elastic string of natural length 50 cm and modulus 20 N has one end fastened to O on a smooth horizontal table. The other end has a particle of mass 0.4 kg attached to it and the particle is released from rest at a distance 60 cm from O. Find the time it takes to reach O.
While the string is stretched
T=kx/l
=20x/0.5
=40x
40x=-.4a
So,a = -100x
w=10
T=2pi/10 =pi/5
I get stuck from here. Can anyone help please? Thanks

2. It would help if you said what "T", "w", etc. mean! I am going to guess that it is the period for the motion-the time required for the string to move from "all the way up" to "all the way down", then back to its starting point. The time required to go from "all the way up" to the central location is 1/4 of T.

3. Originally Posted by HallsofIvy
It would help if you said what "T", "w", etc. mean! I am going to guess that it is the period for the motion-the time required for the string to move from "all the way up" to "all the way down", then back to its starting point. The time required to go from "all the way up" to the central location is 1/4 of T.
Sincere apologies. 1st T is Tension and the second is the period. w is angular frequency. Thanks. Thats quite clear. How do I proceed from there please? It seems the total time would include the period and the time to the central location. Is that right. Thanks.

4. spring constant,

$k = \frac{20 \, N}{0.5 \, m} = 40 \, N/m$

period,

$T = 2\pi \sqrt{\frac{m}{k}} = \frac{\pi}{5}$

the time from maximum displacement back to equilibrium is 1/4 of the period ...

$t = \frac{\pi}{20} \, s$

5. Originally Posted by skeeter
spring constant,

$k = \frac{20 \, N}{0.5 \, m} = 40 \, N/m$

period,

$T = 2\pi \sqrt{\frac{m}{k}} = \frac{\pi}{5}$

the time from maximum displacement back to equilibrium is 1/4 of the period ...

$t = \frac{\pi}{20} \, s$
Thanks a lot! Sorry for late reply.

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