JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Elasticity & SHM

A wooden block performs SHM on a frictionless surface with frequency,

v_{0}

. The block carries a charge

+ Q

on its surface. If now a uniform electric field

E

is switched-on as shown, then the SHM of the block will be

(A)of the same frequency and with shifted mean position.✓ Correct
(B)of the same frequency and with the same mean position.
(C)of changed frequency and with shifted mean position.
(D)of changed frequency and with the same mean position.

Answer: (A)

Solution

The electric field exerts a constant force

F = QE

on the charged block. This constant force shifts the equilibrium position but does not change the restoring force constant of the spring. Since frequency

ν_{0} = \frac{1}{2 π} \frac{k}{m}

depends only on k and m (both unchanged), the frequency remains the same. The mean position shifts by

x_{0} = \frac{QE}{k}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option

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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Elasticity & SHM

Solution

More JEE Advanced questions on Elasticity & SHM