Oscillations

JEE Main level, numerical answer, medium

115 of 194

Question 115 of 194

A mass mm attached to free end of a spring executes SHM with a period of 1s1s. If the mass is increased by 3kg3kg the period of oscillation increases by one second, the value of mass mm is ________ kgkg.

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Answer: 1

Time period of oscillation is given by T=2πmkT=2\pi \sqrt{\dfrac{m}{k}}, where kk is the force constant of the spring.
∵∵ Spring is same, therefore, spring constant will be same.
Putting the different values,
In the first case T=1=2πmkT=1=2\pi \sqrt{\dfrac{m}{k}}
In the second case T′=2=2πm+3k{T}^{'}=2=2\pi \sqrt{\dfrac{m+3}{k}}
Dividing the equations, we get
TT′=12=mm+3\dfrac{T}{{T}^{'}}=\dfrac{1}{2}=\sqrt{\dfrac{m}{m+3}}
⇒mm+3=14⇒4m=m+3⇒m=1\Rightarrow \dfrac{m}{m+3}=\dfrac{1}{4} \Rightarrow 4m=m+3 \Rightarrow m=1
Thus, m=1kgm=1kg

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