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13.Oscillations
medium
A body of mass $5\; kg$ hangs from a spring and oscillates with a time period of $2\pi $ seconds. If the ball is removed, the length of the spring will decrease by
A
$g/k\;metres$
B
$g\;metres$
C
$k/g\;metres$
D
$2\pi\;metres$
(AIPMT-1994)
Solution
Mass $(m)=5 kg$ and
time period $(T)=2 \pi$ sec.
Therefore time period $T=2 \pi \times \sqrt{\frac{m}{k}} \Rightarrow \sqrt{\frac{5}{k}}=1$
or $k=5 N / m$.
According to Hooke's Law, $F=-k l$.
Therefore decrease in length $(l)=-\frac{F}{k}=-\frac{5 g}{5}$ $=-g$ metres
Standard 11
Physics