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BITSAT 2015 Paper

Section: Physics

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Question : 24 of 150

Marks: +1, -0

A load of mass m falls from a height h on to the scale pan hung from the spring as shown in the figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to the pan, then the amplitude of vibration is

mg/d
$\frac{m g}{k} \sqrt{ \frac{1+2 h k}{m g} }$
$\frac{m g}{k} + \frac{m g}{k} \sqrt{ \frac{1+2 h k}{m g} }$
$\frac{m g}{k} \sqrt{ \frac{1+2 h k}{m g} - \frac{m g}{k} }$

Solution:

According to energy conservation principle,

If,

x_1

is maximum elongation in the spring when the particle is in its lowest extreme position. Then,

mgh =

\frac{1}{2} k x_1^2

-

m g x_1

⇒

\frac{1}{2} k x_1^2 - m g x_1

- mgh = 0

or

x_1^2

-

\frac{2 m g}{k} x_1 - \frac{2 m g}{k}

h = 0

∴

x_1

=

\frac{ \frac{2 m g}{k} \pm \sqrt{ \left( \frac{2 m g}{k} \right)^2 + 4 \times \frac{2 m g}{k} h } }{2}

Amplitude A =

X_1 - X_0

(elongation in spring for equilibrium position)

A =

\frac{m g}{k}

\sqrt{ \frac{1+2 h k}{m g} }

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