Question

Mass ${}^{\prime }{m}^{\prime }$ is suspended from the massless pulley which itself is supported by two springs connected to each other as shown. If the mass is slightly displaced from its equilibrium position, the frequency of resulting SHM of mass ${}^{\prime }{m}^{\prime }$ is

• A. $\frac{1}{2\pi }\sqrt{\frac{k_1{k}_2}{(k_1+k_2)m}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{2k_1{k}_2}{(k_1+k_2)m}}$
• C. $\frac{1}{4\pi }\sqrt{\frac{2k_1{k}_2}{(k_1+k_2)m}}$
• D. $\frac{1}{\pi }\sqrt{\frac{k_1{k}_2}{(k_1+k_2)m}}$

Answer 1

The correct option is D $\frac{1}{\pi }\sqrt{\frac{k_1{k}_2}{(k_1+k_2)m}}$

Let consider mass $\left(m\right)$ be displaced by a distance of $x$ from its equilibrium positions. So, that the addition elongation occurs in both springs
$x_1,x_2$ be the addition alongation of spring 1 and spring to respectively.
$2x=x_1+x_2\dots \left(i\right)$
Since the tension of both springs are same.
$T=x_1{x}_2=k_2{x}_2\dots \left(ii\right)$
From equation $\left(ii\right)$,
$x_2=x_1\left(\frac{k_1}{k_2}\right)$
Substitute this value in equation $\left(i\right)$,
$2x=x_1+x_1\left(\frac{k_1}{k_2}\right)$
$2x=x_1(1+\frac{k_1}{k_2})$
$x1=\frac{2x}{(1+\frac{k_1}{k_2})}$
Use this value of $x_1$, in equation $\left(ii\right)$,
$T=k_{1}x1=\frac{k_{1}2x}{1+\frac{k_1}{k_2}}$
$T=\frac{2k_1{k}_2}{k_1+k_2}\dots \left(iii\right)$
Now, $T^{\prime }=2T$
$T^{\prime }=2\times +\frac{2k_1{k}_2}{k_1+k_2}$
$T^{\prime }=\frac{4k_1{k}_2}{k_1+k_2}\dots \left(iv\right)$
We know that: $f=\frac{1}{\pi }\sqrt{\frac{K}{m}}$
Since $T\text{and}{T}^{\prime }$ additional tension over and above the tension is already present at the equilibrium.
Hence, for the mass $m,$
$-\frac{m{d}^{2}x}{d{t}^2}=\frac{4k_1{k}_{2}x}{k_1+k_2}$
Frequency can be defined as reciprocal of time period. $f=\frac{1}{T}$
$f=\frac{1}{2\pi }\sqrt{\frac{K}{m}}$
$f=\frac{1}{2\pi }\sqrt{\frac{4k_1{k}_2}{(k_1+k_2)m}}$
$f=\frac{2}{2\pi }\sqrt{\frac{k_1{k}_2}{(k_1+k_2)m}}$
$f=\frac{1}{\pi }\sqrt{\frac{k_1{k}_2}{(k_1+k_2)m}}$
Final answer : (b)