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Find the time period of mass M when displaced from its equilibrium position and then released for the system as shown in the figure.

Answers (1)

If we pull mass M and then release it, it oscillates with the pulley up and down. Let x0 be the extension of the string when loaded with M. Due to acceleration and the same amount of force, the extension and compression of the spring from the initial position are larger and smaller, respectively. Hence, we can neglect the gravitational force here. Now let us apply force ‘F’ to pull M downwards by displacement x. As the string cannot be extended, its extension will be 2x.

Thus, the total extension = $(x_{0} + 2x)$

When we pull it downward by x

$T^{'}=k(x_{0} + 2x)$

When we do not pull M,

$T^{'}=kx_{0}$

F = 2T

Thus, $F=2kx_{0}$

& $F{'} = 2T{'}$

$F{'} = 2k (x_{0} + 2x)$

Now, restoring force,

Frest $= - (F^{'}-F)$

= $-[2k(x_{0} + 2x) - 2kx_{0}]$

= $-2k.2x$

Thus, $Ma = -4kx$

$a = -\frac{4k}{M }x$

Thus, $a \propto -x$

Therefore, it is a simple harmonic motion.

Now, $a = -\omega ^{2}x$

$\omega ^{2}=- \frac{a}{x}$

$= \frac{+4k}{M}$

$\omega = 2\sqrt{\frac{k}{M}}$

$T = \pi \sqrt{\frac{M}{k} }$

Posted by

infoexpert24

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