Question

A body dropped from height $h$ into the floor makes elastic collision with the floor. The frequency of oscillation of its periodic motion is :

• A. $\frac{1}{2}\sqrt{\frac{g}{2h}}$
• B. $\frac{1}{2}\sqrt{\frac{2h}{g}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{g}{2h}}$
• D. $2\pi \sqrt{\frac{g}{2h}}$

Answer 1

Answer: (A)

$\frac{1}{2}\sqrt{\frac{g}{2h}}$

Time period of two consecutive collision = time of flight of Projectile

$s=ut+\frac{1}{2}a{t}^2$

$u=0$

$s=h=\frac{1}{2}a{t}^2$

$t=\sqrt{\frac{2h}{g}}$

Time period, $T=2t=2\sqrt{\frac{2h}{g}}$

Oscillation frequency, $f=\frac{1}{T}=\frac{1}{2}\sqrt{\frac{g}{2h}}$

Hence, oscillation frequency $\frac{1}{2}\sqrt{\frac{g}{2h}}$