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Physics Question on Oscillations

A 10 kg10\,kg metal block is attached to a spring of spring constant 1000 nm−11000 \, nm^{-1}. A block is displaced from equilibrium position by 10 cm10\,cm and released. The maximum acceleration of the block is

A

10 ms−210 \, ms^{-2}

B

100 ms−1100\, ms ^{-1}

C

200 ms−2200 \, ms^{-2}

D

0.1 ms−20.1 \, ms^{-2}

Answer

10 ms−210 \, ms^{-2}

Explanation

Solution

We know that spring does SHM
So, the restoring force is proportional to displacement
i.e., F=−mω2yF=-m \omega^{2} y...(i)
F=−kyF=-k y...(ii)
where k=k= force constant of the spring
m=10 kgm=10\, kg
A=10 cm=0.1 mA=10\, cm =0.1\, m
Comparing the both equation, we get
ω2=km\omega^{2}=\frac{k}{m}
⇒ω=km=100010\Rightarrow \omega =\sqrt{\frac{k}{m}}= \sqrt{\frac{1000}{10}}
=10 rad/s=10\, rad / s
and acceleration in SHM
amax⁡=−ω2⋅ya_{\max }=-\omega^{2} \cdot y
where ω2\omega^{2} is constant
=−102×(0.1)=−10m/s2=10 ms−2=-10^{2} \times(0.1)=-10 m / s ^{2}=10\, ms ^{-2}