Question

The angular velocities of three bodies in simple harmonic motion are $$\omega_1, \omega_2, \omega_3$$ with their respective amplitudes as $$A_1, A_2, A_3$$. If all the three bodies have same mass and velocity, the.

A
A21ω21=A22ω22=A23ω2
B
A21ω1=A22ω2=A23ω3
C
A1ω21=A2ω22=A3ω23
D
A1ω1=A2ω2=A3ω3

Solution

The correct option is D $$A_1\omega_1=A_2\omega_2=A_3\omega_3$$Rate of change of displacement is known as velocity. The equation for displacement (y) of a body in SHM with angular velocity $$\omega$$ is given by$$y=a\sin\omega t$$where a is amplitudevelocity v$$=$$ rate of change of displacement $$=\displaystyle\frac{dy}{dt}$$$$v=\displaystyle\frac{dy}{dt}=\frac{d}{dt}(a\sin\omega t)=a\omega \cos \omega t$$Given, $$v_1=v_2=v_3$$ and $$a_1=A_1, a_2=A_2$$.$$a_2=A_3$$$$A_1\omega_1 = A_2\omega_2 = A_3\omega_3$$.Physics

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