Question

A smooth wedge of mass m and angle of inclination 60 degrees rests unattached between two springs of spring constant k and 4k, on a smooth horizontal plane, both springs in the un-extended position. The time period of small oscillations of the wedge (assuming that the springs are constrained to get compressed along their length) equals

• A. $\pi (1+\frac{1}{2})\sqrt{\frac{m}{k}}$
• B. $\pi (1+\frac{1}{\sqrt{3}})\sqrt{\frac{m}{k}}$
• C. $\pi (1+\frac{1}{\sqrt{3}})\sqrt{\frac{m}{k}}$
• D. None of above

Answer 1

The correct option is B $\pi (1+\frac{1}{\sqrt{3}})\sqrt{\frac{m}{k}}$

If wedge moved left by x spring will be compressed by y

Horizontal force on wedge = Fx =- ky sin 60$\circ$ = ma

= $-4k\times (sin60{)}^2=ma$

= $-4k\times \frac{3}{4}=ma$

$a=\frac{-3k}{mx}$

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{3k}}$

$t_1=\frac{T}{2}=\pi \sqrt{\frac{m}{3k}}$

Since the wedge is in contact with the spring for half the time period

Similarly for the right spring

Time period will be

$T=2\pi \sqrt{\frac{m}{k}}$

$t_2=\frac{T}{2}=\pi \sqrt{\frac{m}{k}}$

total tiem of oscillation= $t_1+t_1=\pi \sqrt{\frac{m}{k}}(1+\frac{1}{\sqrt{3}})$

option B is correct