Physics - SHM Question with Solution | TestHub

(A)

For equilibrium:

$2mg=kx$

$x=\frac{2mg}{k}$

$A=\frac{2mg}{k}$

$v_{max}=A\omega$

$=(\frac{2mg}{k})\sqrt{\frac{k}{m}}$

$=2g\sqrt{\frac{m}{k}}$

$a_{max}={\omega }^{2}A$

$=(\frac{k}{m})\times (\frac{2mg}{k})$

$=2g$

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(B)

Same as (A)

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(C)

Initially:

$mg=k{x}_0$

$3mg=kx$

$A=\frac{2mg}{k}$

Spring will not acquire natural length.

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(D)

$mg=k{x}_0$

Spring is compressed by $\frac{mg}{k}$ in equilibrium.

For the block:

$kx=mg$ (upward)

Weight = $mg$ (downward)

Applied force $F=2mg$ (upward)

So,

$x=\frac{mg}{k}$

Amplitude:

$A=\frac{2mg}{k}$

Velocity at natural length:

$v=\omega \sqrt{A^2-x^2}$

where

$x=\frac{mg}{k}$