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Byju's Answer

Standard XII

Physics

Frequency to Angular Frequency

The spring bl...

Question

The spring block system as shown in the figure is in equilibrium. The string connecting blocks A and B is cut. The mass of all the three blocks is m and the spring constant of both the springs is k. The amplitude of resulting oscillation of block A is

\frac{m g}{k}

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\frac{2 m g}{k}

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\frac{3 m g}{k}

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\frac{4 m g}{k}

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Solution

The correct option is B $\frac{2 m g}{k}$

Let $x$ is the extension of spring just after cutting.
applying Hooke's law, $m g + m g + m g = k x, x = \frac{3 m g}{k}$
The extension in the spring when the block A in the mean position $= \frac{m g}{k}$
Amplitude of oscillation of $A = \frac{3 m g}{k} - \frac{m g}{k} = \frac{2 m g}{k}$

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The spring block system as shown in the figure is in equilibrium. The string connecting blocks A and B is cut. The mass of all the three blocks is m and the spring constant of both the springs is k. The amplitude of resulting oscillation of block A is

The correct option is B 2mgk Let x is the extension of spring just after cutting.applying Hooke's law, mg+mg+mg=kx,x=3mgkThe extension in the spring when the block A in the mean position =mgkAmplitude of oscillation of A=3mgk−mgk=2mgk