Question

A particle of mass $m$ is attached to three identical springs $\text{A, B}$ and $\text{C}$ each of force constant $k$ as shown in figure. If the particle of mass $m$ is pushed slightly against the spring $\text{A}$ and released, then the time period of oscillation is

• A. $2\pi \sqrt{\frac{2m}{k}}$
• B. $2\pi \sqrt{\frac{m}{2k}}$
• C. $2\pi \sqrt{\frac{m}{k}}$
• D. $2\pi \sqrt{\frac{m}{3k}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{m}{2k}}$

Let the mass is intially at point $\text{O}$. When the mass is pushed against spring $\text{A}$ and released then the point $\text{O}$ is shifted to $\text{O'}$ by $x$ as shown in figure below.


Since, ${\text{OO}}^{{}^{\prime }}=x$

Then $\text{O'M}=\text{O'N}\approx \frac{x}{\sqrt{2}}$
i.e. elongation in spring $\text{B}$ and $\text{C}$ is $\frac{x}{\sqrt{2}}$, while compression in spring $\text{A}$ is $x$.

Taking component of forces in the spring along the direction in which spring $A$ produces its restoring force we get,

Net restoring force $F$ acting on the system,

$F=-[kx+\frac{kx}{\sqrt{2}}\cos {45}^{\circ }+\frac{kx}{\sqrt{2}}\cos {45}^{\circ }]$

$\Rightarrow F=-[kx+\frac{2kx}{\sqrt{2}}\cos {45}^{\circ }]$

$\Rightarrow F=-2kx$

So, the acceleration of the system,
$\therefore a=\frac{F}{m}=-\frac{2k}{m}x$

Now, time period $\left(T\right)$ of the oscillation
$T=2\pi \sqrt{\left|\frac{x}{a}\right|}$

Substituting the value of $a$,
$\therefore T=2\pi \sqrt{\frac{m}{2k}}$

Hence, option (b) is correct answer.