Question

A particle executes SHM of period $\pi$ and amplitude 2cm. Find the acceleration of particle when it is 1 cm from mean position.

• A. $3cm{s}^{-2}$
• B. $4cm{s}^2$
• C. $5cm{s}^{-2}$
• D. $1cm{s}^2$

Answer 1

The correct option is B

$4cm{s}^2$

$x=Asin(\omega t+{\varphi }_0)isthegenerale{q}^{n}forSHMhereA=2cm$

$T=\pi s,\omega =\omega =\frac{2\pi }{T}=2s^7$

$x=2sin(2t+{\varphi }_0)$.....(1)

$v=\frac{dx}{dt}=4cos(2t+{\varphi }_0)$....(2)

$v=\frac{d^{2}x}{d{t}^2}=8sin(2t+{\varphi }_0)$....(3)

Given that $x=1cm,from\left(1\right)sin(2t+{\varphi }_0)=\frac{1}{2}$

from (3) $a=-8\times \frac{1}{2}=-4m{s}^{-1}$.