JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Elasticity & SHM

WhatsApp Email

In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is:

(A) $\frac{1}{2 π} \frac{k}{2 m}$
(B) $\frac{1}{2 π} \frac{k}{Mg s i n α}$
(C) $\frac{1}{2 π} \frac{2 k}{M}$ ✓ Correct
(D) $\frac{1}{2 π} \frac{2 k}{Mg s i n α}$

Answer: (C)

Solution

The component

M g sin α

of the weight along the incline is already balanced by the springs at the equilibrium position, so it only fixes where that position lies and drops out of the equation of motion. For a displacement

x

from equilibrium along the incline, one spring is compressed by

x

while the other is stretched by

x

, and both act back towards equilibrium:

F = - (k x + k x) = - 2 k x

The effective stiffness is

2 k

, so

ω = \frac{2 k}{M}

and the frequency of oscillation is

f = \frac{1}{2 π} \frac{2 k}{M}

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Elasticity & SHM →

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Elasticity & SHM

Solution

More JEE Main questions on Elasticity & SHM