On a smooth inclined plane,a block of mass M | vedclass

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(A) When the block of mass $M$ is displaced by a small distance $x$ along the inclined plane,one spring gets compressed by $x$ and the other gets stre

Vedclass · Accepted Answer

$2 \pi\left(\frac{M}{2 k}\right)^{1 / 2}$

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(A) When the block of mass $M$ is displaced by a small distance $x$ along the inclined plane,one spring gets compressed by $x$ and the other gets stretched by $x$.The restoring force exerted by each spring is $F = -kx$.Since both springs act in the same direction to restore the equilibrium position,the total restoring force is $F_{net} = -kx - kx = -2kx$.The equation of motion for the block is $M a = -2kx$,which can be written as $M \frac{d^2x}{dt^2} + 2kx = 0$.This is the standard form of the simple harmonic motion equation $\frac{d^2x}{dt^2} + \omega^2 x = 0$,where $\omega^2 = \frac{2k}{M}$.The angular frequency is $\omega = \sqrt{\frac{2k}{M}}$.The time period of oscillation is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{M}{2k}}$.Thus,the correct option is $A$.

On a smooth inclined plane,a block of mass $M$ is fixed to two rigid supports using two springs,as shown in the figure. If each spring has spring constant $k$,then the period of oscillation of the block is (Neglect the masses of the springs)

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