Question

Four massless springs whose force constants are $2k,2k,k$ and $2k$, respectively, are attached to a mass $M$ kept on a frictionless horizontal plane (as shown in figure). If the other ends of the springs are fixed and the mass $M$ is displaced , then the frequency of oscillation of the system is

• A. $\frac{1}{2\pi }\sqrt{\frac{k}{4M}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{4k}{M}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{k}{7M}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{7k}{M}}$

Answer 1

The correct option is B $\frac{1}{2\pi }\sqrt{\frac{4k}{M}}$

The two springs on left side to the mass $M$ are having spring constant of $2k$ each. From the figure we can say that they are in series combination;
Equivalent spring constant is $k_{eq}^{\prime }=\frac{k_1{k}_2}{k_1+k_2}=\frac{2k\left(2k\right)}{2k+2k}=k..........\left(1\right)$.
The two springs on right hand side of mass $M$ are in parallel combination. Their effective spring constant is $k_{eq}^{\prime \prime }=(k+2k)=3k......\left(2\right)$.
From $\left(1\right)$ and $\left(2\right)$
Equivalent spring constants of value $k$ and $3k$ are in parallel combination.
Thus, the effective spring constant of the system of springs is given by $k_{eq}=k_{eq}^{\prime }+k_{eq}^{\prime \prime }=k+3k=4k$
Angular frequency of the oscillation of mass $M$ is given by $\omega =\sqrt{\frac{k_{eq}}{M}}=\sqrt{\frac{4k}{M}}$
But ,
$\omega =2\pi f=\sqrt{\frac{4k}{M}}\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{4k}{M}}$
Thus, option (b) is the correct answer.