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What is the period of small oscillations of the block of mass m, if the springs are ideal and pulleys are massless?
                                

A) $\dfrac{{{\pi }}}{{{2}}}\sqrt {\dfrac{{{m}}}{{{k}}}} $
B) $\dfrac{{{\pi }}}{{{2}}}\sqrt {\dfrac{{{m}}}{{{{2k}}}}} $
C) $\dfrac{{{\pi }}}{{{2}}}\sqrt {\dfrac{{{{2m}}}}{{{k}}}} $
D) ${{\pi }}\sqrt {\dfrac{{{m}}}{{{k}}}} $

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Last updated date: 20th Jun 2024
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Answer
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Hint: First find the spring constant and then by using the equation that gives the time period of oscillation of a spring in relation to mass of the body and the spring constant find the weight of the body. Here first we will draw the free body diagram of the given system as shown below. And then after balancing the force as discussed below.

Complete step by step solution:
                                        

Free body diagram of given pulley-block system as shown in below figure.
                                           

$ \Rightarrow $ $2T$ = $mg$
By using Hooke's law for a spring balance.
$ \Rightarrow $ $mg$ =${{4k}}{{{x}}_{{0}}}$ ………….. (1)
$ \Rightarrow $ $\dfrac{{{T}}}{{{2}}} = {{k}}{{{x}}_{{0}}}$
$ \Rightarrow $ ${{T = 2k}}{{{x}}_{{0}}}$
If displaced
$ \Rightarrow$ $mg−2T$ = $ma$
$ \Rightarrow $ $mg−4k({x_0} + x)$ = $ma$
$ \Rightarrow $ $\dfrac{{{T}}}{{{2}}}$= $k({x_0}+x)$
$ \Rightarrow $ $T$ = $2k({x_0}+x)$
Now,
From equation (1), we will get
$ \Rightarrow $ $mg$ =${{4k}}{{{x}}_{{0}}}$
$ \Rightarrow $ ${{{x}}_{{0}}}{{ = }}\dfrac{{{{mg}}}}{{{{4k}}}}$
As we know,
$ \Rightarrow $${{\omega = }}\sqrt {\dfrac{{{g}}}{{{{{x}}_{{0}}}}}} $
Put ${{{x}}_{{0}}}{{ = }}\dfrac{{{{mg}}}}{{{{4k}}}}$and get
$ \Rightarrow $${{\omega = }}\sqrt {\dfrac{{{g}}}{{\dfrac{{{{mg}}}}{{{{4k}}}}}}} $
$ \Rightarrow $${{\omega = }}\sqrt {\dfrac{{{{4k}}}}{{{m}}}} $
Now, by using the formula of time period as ${{T = }}\dfrac{{{{2\pi }}}}{{{\omega }}}$
Put the value of ${{\omega = }}\sqrt {\dfrac{{{{4k}}}}{{{m}}}} $and we will get
$ \Rightarrow $${{T = 2\pi }}\sqrt {\dfrac{{{m}}}{{{{4k}}}}} $
$ \therefore $${{T = \pi }}\sqrt {\dfrac{{{m}}}{{{k}}}} $
Thus, ${{T = \pi }}\sqrt {\dfrac{{{m}}}{{{k}}}} $ is the period of small oscillations of the block of mass m.

Therefore, option (D) is the correct option.

Note: Mathematically, Hooke's law for a spring balance can be written down as:
F= -kx
Where F is the restoring force,
k is the spring constant of the spring balance and
x is the displacement from the initial position of the spring balance system.
This formula is extremely useful in other chapters also like Simple Harmonic Chapter, Newton's Laws of Motion, etc.