Question

Two identical springs of spring constant $k$ are attached to a block of mass $m$ and to fixed supports as shown in the figure. The time period of oscillation is

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

Answer 1

Answer: (B)

Let the mass $m$ be displaced by a small distance $x$ to the right from its mean position as shown in figure $(b)$. Due to it the spring on the left side gets stretched by a length $x$ while that on the right side gets compressed by the same length. The forces acting on the mass are
$F_1=-kx$ towards left hand side
$F_2=-kx$ towards left hand side
The net force acting on the mass is, $F=F_1+F_2=-2kx$
Here, $F\propto x$ and $-ve$ sign shows that force is towards the mean position, therefore the motion executed by the particle is simple harmonic. Its acceleration is
$a=\frac{F}{m}=-\frac{2kx}{m}...\left(i\right)$
The standard equation of $SHM$ is
$a=-{\omega }^{2}x...\left(ii\right)$
Comparing $\left(i\right)$ and $\left(ii\right)$, we get
${\omega }^2=\frac{2k}{m}$ or $\omega =\sqrt{\frac{2k}{m}}$
Time period, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{2k}}$