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Simple Harmonic Motion question

2011 · Shift 2 · Q42
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Simple Harmonic Motion question

2011 · Shift 2 · Q42

JEE AdvancedPhysicsSimple Harmonic MotionMCQ+2 / −0.5
A wooden block performs SHMSHMSHM on a frictionless surface with frequency, v0.{v_0}.v0​. The block carries a charge +Q+Q+Q on its surface . If now a uniform electric field E→\overrightarrow EE is switched- on as shown, then the SHMSHMSHM of the block will be IIT-JEE 2011 Paper 2 Offline Physics - Simple Harmonic Motion Question 20 English
  1. A
    of the same frequency and with shifted mean position.
  2. B
    of the same frequency and with the same mean position
  3. C
    of changed frequency and with shifted mean position.
  4. D
    of changed frequency and with the same mean position.
View written solutionFree

Correct answer: A

  1. Initial SHM setup

A block attached to a spring on a frictionless surface performs SHM with frequency ν0\nu_0ν0​.

For a spring-mass system,

ν0=12πkm\nu_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}ν0​=2π1​mk​​

where kkk is the spring constant and mmm is the mass of the block.

  1. Effect of switching on the uniform electric field

The block has charge +Q+Q+Q, so in a uniform electric field E⃗\vec EE, it experiences a constant electric force

Fe=QEF_e = QEFe​=QE

(along the direction of the field).

Since this force is constant, it does not depend on displacement xxx.

  1. New equation of motion

Let xxx be measured from the old mean position. Then the forces are:

  • Spring restoring force: −kx-kx−kx
  • Electric force: +QE+QE+QE (taking field direction as positive)

So,

mx¨=−kx+QEm\ddot x = -kx + QEmx¨=−kx+QE

Rearranging,

mx¨+kx=QEm\ddot x + kx = QEmx¨+kx=QE
  1. Finding the new mean position

At the new equilibrium position, acceleration is zero:

0+kx0=QE0 + kx_0 = QE0+kx0​=QE

so,

x0=QEkx_0 = \frac{QE}{k}x0​=kQE​

Thus the mean position shifts by

QEk\frac{QE}{k}kQE​

in the direction of the electric field.

  1. Frequency about the new mean position

Let

y=x−x0y = x - x_0y=x−x0​

be displacement from the new equilibrium position. Then the equation becomes

my¨+ky=0m\ddot y + ky = 0my¨​+ky=0

which is again standard SHM.

Therefore the angular frequency remains

ω=km\omega = \sqrt{\frac{k}{m}}ω=mk​​

and the frequency remains

ν=12πkm=ν0\nu = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \nu_0ν=2π1​mk​​=ν0​
  1. Conclusion
  • The frequency remains unchanged.
  • The mean position shifts.

So the correct option is:

A\boxed{\text{A}}A​
  1. Comparison with stored answer

Stored correct answer: AAA

My derived answer is also AAA, so they agree.

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