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Simple Harmonic Motion question

2011 · Shift 0 · Q67
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Simple Harmonic Motion question

2011 · Shift 0 · Q67

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
A mass M,M,M, attached to a horizontal spring, executes S.H.M.S.H.M.S.H.M. with amplitude A1.{A_1}.A1​. When the mass MMM passes through its mean position then a smaller mass mmm is placed over it and both of them move together with amplitude A2.{A_2}.A2​. The ratio of (A1A2)\left( {{{{A_1}} \over {{A_2}}}} \right)(A2​A1​​) is :
  1. A
    M+mM{{M + m} \over M}MM+m​
  2. B
    (MM+m)12{\left( {{M \over {M + m}}} \right)^{{1 \over 2}}}(M+mM​)21​
  3. C
    (M+mM)12{\left( {{{M + m} \over M}} \right)^{{1 \over 2}}}(MM+m​)21​
  4. D
    MM+m{M \over {M + m}}M+mM​
View written solutionFree

Correct answer: C

  1. Initial SHM of mass MMM

For the spring-mass system,

ω1=kM\omega_1=\sqrt{\frac{k}{M}}ω1​=Mk​​

with amplitude A1A_1A1​.

When the particle passes through the mean position, its speed is maximum:

v1=ω1A1=A1kMv_1=\omega_1 A_1= A_1\sqrt{\frac{k}{M}}v1​=ω1​A1​=A1​Mk​​
  1. A smaller mass mmm is placed on it at mean position

At that instant, the spring is at natural/mean position, so displacement is zero.

The mass mmm is gently placed and both move together afterward as a single mass M+mM+mM+m.

Since the placing happens instantaneously, we conserve linear momentum at that instant:

Mv1=(M+m)v2Mv_1=(M+m)v_2Mv1​=(M+m)v2​

So,

v2=MM+mv1v_2=\frac{M}{M+m}v_1v2​=M+mM​v1​
  1. New angular frequency

Now the combined mass is M+mM+mM+m, so the new angular frequency is

ω2=kM+m\omega_2=\sqrt{\frac{k}{M+m}}ω2​=M+mk​​
  1. Find new amplitude A2A_2A2​

Right after placing the mass, the system is still at mean position (x=0x=0x=0), and has speed v2v_2v2​.

For SHM, if a particle starts from mean position with speed vmax⁡v_{\max}vmax​, then amplitude is

A=vmax⁡ωA=\frac{v_{\max}}{\omega}A=ωvmax​​

Hence,

A2=v2ω2A_2=\frac{v_2}{\omega_2}A2​=ω2​v2​​

Substitute v2v_2v2​ and ω2\omega_2ω2​:

A2=MM+mv1kM+mA_2=\frac{\frac{M}{M+m}v_1}{\sqrt{\frac{k}{M+m}}}A2​=M+mk​​M+mM​v1​​

Now use

v1=A1kMv_1=A_1\sqrt{\frac{k}{M}}v1​=A1​Mk​​

So,

A2=MM+m⋅A1kM⋅M+mkA_2=\frac{M}{M+m} \cdot A_1\sqrt{\frac{k}{M}} \cdot \sqrt{\frac{M+m}{k}}A2​=M+mM​⋅A1​Mk​​⋅kM+m​​ A2=A1MM+mM+mMA_2=A_1\frac{M}{M+m}\sqrt{\frac{M+m}{M}}A2​=A1​M+mM​MM+m​​ A2=A1MM+mA_2=A_1\sqrt{\frac{M}{M+m}}A2​=A1​M+mM​​

Therefore,

A1A2=M+mM\frac{A_1}{A_2}=\sqrt{\frac{M+m}{M}}A2​A1​​=MM+m​​
  1. Match with options
A1A2=(M+mM)1/2\frac{A_1}{A_2}=\left(\frac{M+m}{M}\right)^{1/2}A2​A1​​=(MM+m​)1/2

This corresponds to Option C.

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