A 20 gm particle is subjected to two simple harmonic motions x_1 = 2 sin 10 t, where x_1& x_2 are in metre& t is in sec.

JEE MainPhysicsModerateSimple Harmonic Motion

A 20 gm particle is subjected to two simple harmonic motions
x_1 = 2 sin 10 t,
$x_{2} = 4sin\left( 10t + \frac{\pi}{3} \right)$ where x_1& x_2 are in
metre& t is in sec.

Show answer and solution

At t = 0
Displacement

x = x_{1} + x_{2} = 4sin\frac{\pi}{2} = 2\sqrt{3}m

Resulting Amplitude

A = \sqrt{2^{2} + 4^{2} + 2(2)(4)cos\pi/3} = \sqrt{4 + 16 + 8} = \sqrt{28} = 2\sqrt{7}m

Maximum speed

= A\omega = 20\sqrt{7}m/s

Maximum acceleration

= A\omega^{2} = 200\sqrt{7}m/s^{2}

Source: Grass-G / P-23 Physcis Paper 31 Dec FST.docx

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