Question

A system is shown in the figure. The time period for small oscillations of the two identical blocks will be.

• A. $2\pi \sqrt{\frac{3m}{k}}$
• B. $2\pi \sqrt{\frac{3m}{2k}}$
• C. $2\pi \sqrt{\frac{3m}{4k}}$
• D. $2\pi \sqrt{\frac{3m}{8k}}$

Answer 1

Answer: (B)

$2\pi \sqrt{\frac{3m}{2k}}$

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
$a=\frac{F}{m}=-\frac{2kx}{m}$ ....(i)
The standard equation of SHM is
$a=-{\omega }^{2}x$ ...(ii)
Comparing (i) and (ii), we get
${\omega }^2=\frac{2k}{m}or\omega =\sqrt{\frac{2k}{m}}$

Time period, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{2k}}$