Question

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Answer 1

Given:

If a body of mass m₁ is suspended from a light spring, then

Period of oscillation of body= T

If a body of mass m₂ is tied to the first body and system is made to oscillate, then

Period of oscillation of system= 2T

To Find:

Relation between the masses m₁ and m₂

Solution:

We know that,

• If a body of mass m is suspended from the spring of force constant k, then time period of oscillations of the body is given by

$\boxed{\bf\pink{T=2\pi\sqrt{\frac{m}{k}}}}$

Now, for body of mass m₁

$\sf{Time\:period\:of\:Oscillation=2\pi\sqrt{\dfrac{m_{1}}{k}}}$

$\sf{T=2\pi\sqrt{\dfrac{m_{1}}{k}}}---------(1)$

Now, for a system having bodies of mass m₁ and m₂

$\sf{Time\:period\:of\:Oscillation=2\pi\sqrt{\dfrac{m_{1}+m_{2}}{k}}}$

$\sf{2T=2\pi\sqrt{\dfrac{m_{1}+m_{2}}{k}}}$

From (1), we get

$\sf{2\bigg(2\pi\sqrt{\dfrac{m_{1}}{k}}\bigg)=2\pi\sqrt{\dfrac{m_{1}+m_{2}}{k}}}$

$\sf{\cancel{4\pi}\sqrt{\dfrac{m_{1}}{k}}=\cancel{2\pi}\sqrt{\dfrac{m_{1}+m_{2}}{k}}}$

$\sf{2\sqrt{\dfrac{m_{1}}{k}}=\sqrt{\dfrac{m_{1}+m_{2}}{k}}}$

On squaring both the sides, we get

$\sf{\bigg(2\sqrt{\dfrac{m_{1}}{k}}\bigg)^{2}=\bigg(\sqrt{\dfrac{m_{1}+m_{2}}{k}}\bigg)^{2}}$

$\sf{\dfrac{4m_{1}}{\cancel{k}}=\dfrac{m_{1}+m_{2}}{\cancel{k}}}$

$\sf{4m_{1}=m_{1}+m_{2}}$

$\sf{4m_{1}-m_{1}=m_{2}}$

$\sf{3m_{1}=m_{2}}$

$\green{\boxed{\bf{\sf{m_{1}=\dfrac{1}{3}m_{2}}}}}$

So, mass m₁ is 1/3 times of mass m₂.