Question

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

• A. $1000 \,g$
• B. $10^{5} g$
• C. $10^{7} g$
• D. $10^{4} g$

Answer 1

Answer: (D)

At point of equilibrium $k x=m g$
$k \times 9.8 \times 10^{-3}=100 \times 10^{-3} \times 9.8$
$k=100 \times 10^{-1}$
$k=10 \,N / m$
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$