# Hooke's Law/SHM QuestionWatch

#1
Hi everyone, I've managed to get as far as part d on this, but don't know how to start (answer to part c is 1.21ms-1). I also don't understand the general movement of the particle - it's displaced, then undergoes SHM then suddenly the particle moves up enough to go slack?

0
4 weeks ago
#2
(Original post by RussianQuestion)
Hi everyone, I've managed to get as far as part d on this, but don't know how to start (answer to part c is 1.21ms-1). I also don't understand the general movement of the particle - it's displaced, then undergoes SHM then suddenly the particle moves up enough to go slack?
For part (d) the motion of the particle is in two parts. You have to calculate the time for each part separately then add the times together. The first part of the motion is SHM in the upwards direction from point B (0.95m below A) to the point where the string goes slack (0.8m below A). The particle is still moving upwards at this point and becomes a projectile moving vertically under the influence of gravity. The particle will come to instantaneous rest at the top of its flight as a projectile. You can use SUVAT to determine the time taken from 0.8m below A to the top of the particle's flight.
1
#3
(Original post by old_engineer)
For part (d) the motion of the particle is in two parts. You have to calculate the time for each part separately then add the times together. The first part of the motion is SHM in the upwards direction from point B (0.95m below A) to the point where the string goes slack (0.8m below A). The particle is still moving upwards at this point and becomes a projectile moving vertically under the influence of gravity. The particle will come to instantaneous rest at the top of its flight as a projectile. You can use SUVAT to determine the time taken from 0.8m below A to the top of the particle's flight.
Thank you for your help - as for the calculation, I think I will be fine there. However, does the particle actually oscillate then? Or does it go straight up from 0.95 below A (SHM but not actually passing the equilibrium point more than once) until the string goes slack? In which case, if there's no actual oscillation, how does this count as SHM?
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4 weeks ago
#4
(Original post by RussianQuestion)
Thank you for your help - as for the calculation, I think I will be fine there. However, does the particle actually oscillate then? Or does it go straight up from 0.95 below A (SHM but not actually passing the equilibrium point more than once) until the string goes slack? In which case, if there's no actual oscillation, how does this count as SHM?
Assuming no energy loss, the motion of the particle will be cyclic but not pure SHM. Each complete cycle will be partly SHM (while the string is under tension) and partly projectile-like (while the string is slack).
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