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Question

Physics Question on Oscillations

A body is executing simple harmonic motion. When the displacements from the mean position is 4cm4 \,cm and 5cm5 \,cm, the corresponding velocities of the body is 10cm/sec10 \,cm/sec and 8cm/sec8\, cm/sec. Then the time period of the body is

A

2π2\frac{2\pi}{2} sec

B

π/2\pi / 2 sec

C

π\pi sec

D

(3π/2)( 3 \pi / 2) sec

Answer

π\pi sec

Explanation

Solution

For a body executing SHM, velocity,
v=ω2(a2y2)v=\sqrt{\omega^{2}\left(a^{2}-y^{2}\right)} we have
102=ω2(a242)10^{2}=\omega^{2}\left(a^{2}-4^{2}\right)
and 82=ω2(a252)8^{2}=\omega^{2}\left(a^{2}-5^{2}\right)
So, 10282=ω2(5242)=(3ω)2 10^{2}-8^{2}=\omega^{2}\left(5^{2}-4^{2}\right)=(3 \omega)^{2}
or 6=3ω6=3 \omega
or ω=2 \omega=2
\because Time, t=2π/ωt=2 \pi / \omega
t=2π/2=πsec\therefore t=2 \pi / 2=\pi \,sec