Question

A point mass $m=20 \, kg,$ 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=10 \, m / s^{2}) $

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

Answer 1

Answer: (C)

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