Question

A point mass $m=20kg,$ is suspended by a massless spring of constant 2000 $N/M$ . The point mass is released when elongation in the spring is 15 cm. The equation of displacement of particle as a function of time is (Take $g=10m/s^2)$

• A. $y=10sin10t$
• B. $y=10cos10t$
• C. $y=10sin\left(10t+\frac{\pi }{6}\right)$
• D. None of these

Answer 1

Answer: (C)

The motion of block is S.H.M.
$\therefore$ $y=Asin\left(\omega t+\varphi \right)$
Here amplitude is $A=\frac{mg}{k}$
$=\frac{20\times 10}{2000}m=10cm$
At $t=0,$ displacement of body with respect to mean position is
$y=15-10=5cm$
$\therefore$ $5=10sin\left(\omega \times 0+\varphi \right)$
or $\frac{1}{2}=sin\varphi \Rightarrow \varphi =\frac{\pi }{6}$
$\therefore$ $y=10sin\left(10t+\frac{\pi }{6}\right)$