First slide

Simple hormonic motion

Question

A particle is executing SHM along a straight with an amplitude of 5 cm and time period 2 sec. Find the acceleration of the particle when its velocity is equal to half of its maximum velocity.

Moderate

A

$17.84 cm / s^{2}$
B

$13.59 cm / s^{2}$
C

$42.50 cm / s^{2}$
D

$27.42 cm / s^{2}$

Solution

$Y = ω \sqrt{A^{2} - x^{2}} \Rightarrow x = \sqrt{A^{2} - \frac{V^{2}}{ω^{2}}} = \frac{1}{ω} \sqrt{A^{2} ω^{2} - V^{2}} = \frac{1}{ω} \sqrt{{V_{m}}^{2} - V^{2}} ∴ Accel e ration a = ω^{2} x = ω \sqrt{{V_{m}}^{2} - V^{2}} = ω \sqrt{{V_{m}}^{2} - {(\frac{V_{m}}{2})}^{2}} = \frac{\sqrt{3}}{2} Aω ∴ a = \frac{\sqrt{3}}{2} x 5 x 10^{- 2} x \frac{2 π}{2} m / s^{2} = 13.59 cm / s^{2}$

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