Oscillations

JEE Main level, numerical answer, medium

155 of 194

Question 155 of 194

A rectangular block of mass 5kg5kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 1m1m and time period 3.14s3.14s. The maximum force exerted by spring on block is ___ NN.

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Answer: 20

By Newton's second law,
F=mamaxF=m{a}_{max}.
The maximum acceleration for a SHM is given by
amax=ω2A...(i){a}_{max}={\omega }^{2}A...(i)
The time period is given by
T=2πω=3.14sT=\dfrac{2\pi }{\omega }=3.14s
⇒ω=2πT=2π3.14=2rads−1\Rightarrow \omega =\dfrac{2\pi }{T}=\dfrac{2\pi }{3.14}=2rad{s}^{-1}
Thus, the maximum acceleration from equation (i)
amax=4×1=4ms−2{a}_{max}=4\times 1=4m{s}^{-2}
The maximum force is
F=5kg×4ms−2=20NF=5kg\times 4m{s}^{-2}=20N

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