Question

A point mass $m=20 \, kg,$ is suspended by a massless spring of constant $2000Nm^{- 1}$ . The point mass is released from rest at $t=0$ , when elongation in the spring is 15 cm. The equation of displacement of particle as a function of time is (Take $g=10 \, m / s^{2}\left. \, $

• A. $y=5sin\left(10 t\right)$
• B. $y=5sin \left(10 t + \frac{\pi }{6}\right)$
• C. $y=5cos\left(10 t\right)$
• D. None of these

Answer 1

Answer: (C)

The motion of block is SHM,
$\therefore $ $y=Asin \left(\omega t + \, \phi\right)$
Here, equilibrium or mean position is at elongation,
$x=\frac{m g}{k}$ $=\frac{20 \, \times 10}{2000}m=10 \, cm$
The amplitude of the motion is $A=15cm-10cm=5cm$
At $t=0,$ displacement of body with respect to mean position is
$y=15-10=5cm$
$\therefore $ $5=5sin\left(\omega \times 0 + \phi\right)$
or $1=sin\phi \, \, \Rightarrow \phi=\frac{\pi }{2}$
$\therefore $ $y=5sin\left(10 t + \frac{\pi }{2}\right)=5cos\left(10 t\right)$