Question

A particle in SHM has amplitude 'A' and time
period T. How much time it will take to travel
from x = A to x = A/2
​

Answer 1

Answer:

• T/6 time it will take to travel from x = A to x = A/2.

Explanation:

Given that,

• A particle in Simple harmonic motion (SHM) has amplitude 'A' and time period 'T'.

According to Question,

$\sf \red{ For \: \: SHM, } \: x = Asin(\frac{2π}{T}.t)$

[ At x = A ]

$\leadsto \sf \: A = Asin(\frac{2π}{T}.t) \\ \\ \leadsto \sf \: sin( \frac{2 \pi }{T} .t) = \frac{A}{A} \\ \\ \leadsto \sf \: sin( \frac{2 \pi }{T} .t) =1 \\ \\ \leadsto \sf \: sin( \frac{2 \pi }{T} .t) = sin( \frac{ \pi}{2} ) \\ \\ \leadsto \sf t = \frac{T}{4} \: \red \bigstar$

[ At x = A/2 ]

$\leadsto \sf \: \frac{A}{2} = Asin(\frac{2π}{T}.t) \\ \\ \leadsto \sf \: sin \frac{\pi}{6} = sin( \frac{2\pi}{T} .t) \\ \\ \leadsto \sf \: t = \frac{T}{12} \: \green \bigstar$

$\sf \: \underline{Now, } \\ \\ \sf\leadsto Time \: taken \: to \: travel \: from \: x = A \: to \: x = \frac{A}{2}. \\ \\ \leadsto \sf \: \frac{T}{4} - \frac{T}{12} \\ \\ \leadsto \sf \: \frac{3 \times T }{4 \times 3} - \frac{T}{12} \\ \\ \leadsto \sf \: \frac{3T}{12} - \frac{T}{12} \\ \\ \leadsto \sf \: \frac{3T -T }{12} \\ \\ \leadsto \sf \: \frac{2T}{12} \\ \\ \leadsto \sf \: \frac{T}{6} \: \: \purple\bigstar$