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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
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B
A21ω1=A22ω2=A23ω3
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C
A1ω21=A2ω22=A3ω23
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D
A1ω1=A2ω2=A3ω3
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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 amplitude
velocity 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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