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Simple Harmonic Motion question

2013 · Shift 0 · Q67
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  5. /2013 · Shift 0 · Q67

Simple Harmonic Motion question

2013 · Shift 0 · Q67

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M.M.M. The piston and the cylinder have equal cross sectional area AAA. When the piston is in equilibrium, the volume of the gas is V0{V_0}V0​ and its pressure is P0.{P_0}.P0​. The piston is slightly displaced from the equilibrium position and released,. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frquency
  1. A
    12π AγP0V0M{1 \over {2\pi }}\,{{A\gamma {P_0}} \over {{V_0}M}}2π1​V0​MAγP0​​
  2. B
    12π V0MP0A2γ{1 \over {2\pi }}\,{{{V_0}M{P_0}} \over {{A^2}\gamma }}2π1​A2γV0​MP0​​
  3. C
    12π AγP0V0M{1 \over {2\pi }}\,\sqrt {{{A\gamma {P_0}} \over {{V_0}M}}}2π1​V0​MAγP0​​​
  4. D
    12π MV0AγP0{1 \over {2\pi }}\,\sqrt {{{M{V_0}} \over {A\gamma {P_0}}}}2π1​AγP0​MV0​​​
View written solutionFree

Correct answer: NO CORRECT OPTION AMONG A, B, C, D., IF OPTION C WERE INTENDED, IT IS MISSING A FACTOR OF $\SQRT{A}$; THE CORRECT EXPRESSION IS $\DISPLAYSTYLE \FRAC{1}{2\PI}\SQRT{\FRAC{A^2\GAMMA P_0}{MV_0}}$.

  1. Equilibrium condition

Let the piston be displaced vertically by a small distance xxx from equilibrium.

Since the cylinder has cross-sectional area AAA, the change in gas volume is

ΔV=Ax\Delta V = AxΔV=Ax

(with sign depending on direction; for restoring force we only need the linear relation).

At equilibrium, the gas pressure is P0P_0P0​ and balances the external forces on the piston.

  1. Nature of compression/expansion

The system is said to be completely isolated, so there is no heat exchange with surroundings. Hence the oscillation is adiabatic.

For an adiabatic change of an ideal gas,

PVγ=constantPV^\gamma = \text{constant}PVγ=constant

Taking differential form for small changes,

d(PVγ)=0d(PV^\gamma)=0d(PVγ)=0 VγdP+γPVγ−1dV=0V^\gamma dP + \gamma P V^{\gamma-1} dV = 0VγdP+γPVγ−1dV=0 dP=−γPdVVdP = -\gamma P \frac{dV}{V}dP=−γPVdV​

So near equilibrium,

ΔP=−γP0ΔVV0\Delta P = -\gamma P_0 \frac{\Delta V}{V_0}ΔP=−γP0​V0​ΔV​

Using ΔV=Ax\Delta V = AxΔV=Ax,

ΔP=−γP0AxV0\Delta P = -\gamma P_0 \frac{Ax}{V_0}ΔP=−γP0​V0​Ax​
  1. Restoring force on the piston

The extra force on the piston due to pressure change is

F=A ΔPF = A\,\Delta PF=AΔP

Thus,

F=A(−γP0AxV0)=−A2γP0V0xF = A\left(-\gamma P_0 \frac{Ax}{V_0}\right) = -\frac{A^2\gamma P_0}{V_0}xF=A(−γP0​V0​Ax​)=−V0​A2γP0​​x

This is of the form

F=−kxF=-kxF=−kx

with effective spring constant

k=A2γP0V0k=\frac{A^2\gamma P_0}{V_0}k=V0​A2γP0​​
  1. Angular frequency and frequency

For SHM,

ω=kM\omega = \sqrt{\frac{k}{M}}ω=Mk​​

Therefore,

ω=A2γP0MV0=AγP0MV0\omega = \sqrt{\frac{A^2\gamma P_0}{MV_0}} = A\sqrt{\frac{\gamma P_0}{MV_0}}ω=MV0​A2γP0​​​=AMV0​γP0​​​

Hence the frequency is

f=ω2π=12πA2γP0MV0f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}f=2πω​=2π1​MV0​A2γP0​​​ f=A2πγP0MV0\boxed{f=\frac{A}{2\pi}\sqrt{\frac{\gamma P_0}{MV_0}}}f=2πA​MV0​γP0​​​​

Equivalently,

f=12πA2γP0MV0\boxed{f=\frac{1}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}}f=2π1​MV0​A2γP0​​​​
  1. Compare with options

Given options are:

  • A: 12πAγP0V0M\frac{1}{2\pi}\frac{A\gamma P_0}{V_0M}2π1​V0​MAγP0​​
  • B: 12πV0MP0A2γ\frac{1}{2\pi}\frac{V_0MP_0}{A^2\gamma}2π1​A2γV0​MP0​​
  • C: 12πAγP0V0M\frac{1}{2\pi}\sqrt{\frac{A\gamma P_0}{V_0M}}2π1​V0​MAγP0​​​
  • D: 12πMV0AγP0\frac{1}{2\pi}\sqrt{\frac{MV_0}{A\gamma P_0}}2π1​AγP0​MV0​​​

Our derived result contains A2A^2A2 inside the square root, not AAA. So none of the listed options matches exactly.

Also, dimensional analysis confirms this:

A2P0MV0\frac{A^2 P_0}{MV_0}MV0​A2P0​​

has dimensions of 1/s21/s^21/s2, while

AP0MV0\frac{A P_0}{MV_0}MV0​AP0​​

does not. So option C is dimensionally incorrect.

  1. Conclusion

The correct frequency should be

f=12πA2γP0MV0\boxed{f=\frac{1}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}}f=2π1​MV0​A2γP0​​​​

which is not present among the options. Therefore, the stored answer C\text{C}C is incorrect.

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