13.Oscillations
medium

A $100 \,g$ mass stretches a particular spring by $9.8 \,cm$, when suspended vertically from it. ....... $g$ large a mass must be attached to the spring if the period of vibration is to be $6.28 \,s$.

A

$1000$

B

$10^5$

C

$10^7$

D

$10^4$

Solution

(d)

At point of equilibrium $k x=m g$

$k \times 9.8 \times 10^{-2}=100 \times 10^{-3} \times 9.8$

$k=100 \times 10^{-1}$

$k=10 \,N / m$

Period of vibration needed $=6.28 \,s$

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

$6.28=2 \times 3.14 \sqrt{\frac{m}{10}}$

$1=\frac{m}{10}$

$m=10 \,kg$ or $10^4 \,g$

Standard 11
Physics

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