First slide

Applications of SHM

Question

The displacement of an object attached to a spring and executing simple harmonic motion is given by ${x=2×10}^{-2} cos(πt)m$ . The time at which the maximum speed first occurs is

Moderate

A

0.5 seconds
B

1 second
C

1.5 seconds
D

2 seconds

Solution

Given displacement ${x=2×10}^{-2} cos(πt)$
$d i f f e r e n t i a t e a b o v e e q u a t i o n w i t h r e s p e c t t o t i m e v e l o c i t y = v= \frac{dx}{dt} = - 2 \times 10^{- 2} πsin(πt) F o r t h e f i r s t t i m e w h e n V = V_{m a x} v e l o c i t y i s m a x i m u m w h e n sinπ t = 1 = \sin \frac{π}{2} π t = \frac{π}{2} t = \frac{1}{2} = 0.5 s e c o n d$
***Alternate solution: The given equation shows that the particle is at extreme position .
since $πt = ω \Rightarrow πt = 2 \frac{π}{T} time period is equal to 2 \sec and the particle will take a time equal to \frac{T}{4} = 0.5 \sec$ to reach the mean position where the velocity is maximum.

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