Question

In the arrangement given in figure, if the block of mass m is displaced, the frequency is given by

A. \[n = \dfrac{1}{{2\pi }}\sqrt {\left( {\dfrac{{{k_1} - {k_2}}}{m}} \right)} \]
B. \[n = \dfrac{1}{{2\pi }}\sqrt {\left( {\dfrac{{{k_1} + {k_2}}}{m}} \right)} \]
C. \[n = \dfrac{1}{{2\pi }}\sqrt {\left( {\dfrac{m}{{{k_1} + {k_2}}}} \right)} \]
D. \[n = \dfrac{1}{{2\pi }}\sqrt {\left( {\dfrac{m}{{{k_1} - {k_2}}}} \right)} \]

Answer 1

Hint: To find the frequency consider the small displacement of the block by x and find the net force acting on the block in terms of their springs constant.

Formula used:
\[T = \dfrac{{2\pi }}{\omega },\,n = \dfrac{1}{T}\] and \[\omega = \sqrt {\dfrac{k}{m}} \]
Where, T= Time period of oscillation,\[\omega \] = Angular frequency of oscillation, n = Frequency of oscillation, k= Spring constant and m = Mass suspended to spring

Complete step by step solution:
Given here is a spring block system of two springs A and B of constant \[{k_1}\] and \[{k_2}\]respectively, we have to find the frequency when block m is displaced from its position.

Let the block m be displaced by x towards the right then spring A will pull the block by force \[{k_1}x\] and compression force acting on spring B will be \[{k_2}x\] as shown in the image below,

Image: Force on the spring block system

Then, the net force acting on the block will be
\[{F_{net}} = {k_1}x + {k_2}x \Rightarrow {F_{net}} = - \left( {{k_1} + {k_2}} \right)x\,.......(1)\]

Negative magnitude on the net force in \[{F_{net}} = - \left( {{k_1} + {k_2}} \right)x\] signifies the opposite direction of net force with respect to displacement.

From F= ma, acceleration a of the block can be expressed as,
\[a = \dfrac{{{F_{net}}}}{m} \Rightarrow a = \dfrac{{ - \left( {{k_1} + {k_2}} \right)x}}{m}\,........(2)\]

Comparing equation (2) with we get,
\[{\omega ^2} = \dfrac{{{k_1} + {k_2}}}{m} \Rightarrow \omega = \sqrt {\dfrac{{{k_1} + {k_2}}}{m}} \,.........(3)\]

Then, the frequency of oscillation is given by,
\[n = \dfrac{1}{T} = \dfrac{\omega }{{2\pi }}\,........(4)\]

Substituting the value of from equation (3) in equation (4) frequency will be,
\[n = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{k_1} + {k_2}}}{m}} \,\]

Therefore, option B is the correct option.

Note: If the acceleration of an oscillating body is of the form \[a = - {\omega ^2}x\], then the body is said to be performing simple harmonic motion (S.H.M.)