First slide

Simple harmonic motion

Question

Two particles A and B of equal masses are suspended from two mass less spring of spring constants $K_{1}$ and $K_{2}$ respectively. If the maximum velocities, during oscillations are equal. The ratio of amplitudes of 'A' and 'B'.

Moderate

A

$\sqrt{\frac{K_{1}}{K_{2}}}$
B

$\frac{K_{1}}{K_{2}}$
C

$\sqrt{\frac{K_{2}}{K_{1}}}$
D

$\frac{K_{2}}{K_{1}}$

Solution

maximum velocity is $v_{max} =Aω$ , here A=amplitude, $ω$ =angular frequency,T=time period, m=mass of oscillator,k=force constant
$ω= (\frac{2π}{T}) V_{m a x} = \frac{A (2π)}{2π \sqrt{\frac{m}{k}}} V_{m a x} =A \sqrt{\frac{k}{m}} Hence consider ratio of maximum velocity at A and B \frac{{v_{max}}_{1}}{{v_{max}}_{2}} = \frac{A_{1}}{A_{2}} \sqrt{\frac{K_{1} /m}{K_{2} /m}} {given v}_{max}_{1} = {v_{max}}_{2} \frac{A_{1}}{A_{2}} = \sqrt{\frac{K_{2}}{K_{1}}}$

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