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Question

Physics Question on Oscillations

A particle executing SHMSHM has a maximum speed of 0.5 ms−10.5 \, ms^{ -1} and maximum acceleration of 1.0 ms−21.0 \, ms^{-2} The angular frequency of oscillation is

A

2 rad s−12 \, rad \, s^{-1}

B

0.5 rad s−10.5 \, rad \, s^{-1}

C

2 π rad s−12\,\pi \, rad \, s^{-1}

D

0.5 π rad s−10.5\, \pi \, rad \, s^{-1}

Answer

2 rad s−12 \, rad \, s^{-1}

Explanation

Solution

Key Idea In SHM,
Maximum velocity vmax⁡=ωAv_{\max }=\omega A
Maximum acceleration amax⁡=Aω2a_{\max }=A \omega^{2}
⇒ω=Aω2Aω=amax⁡vmax⁡\Rightarrow \omega =\frac{A \omega^{2}}{A \omega}=\frac{a_{\max }}{v_{\max }}
vmax⁡=0.5 ms−1v_{\max } =0.5 \,ms ^{-1}
amax⁡=1.0 ms−2a_{\max } =1.0 \,ms ^{-2}
ω=10.5=105=2=2 rads−1\omega =\frac{1}{0.5}=\frac{10}{5}=2=2\, rad s ^{-1}