First slide

Introduction to oscillations

Question

A body is dropped on to the floor from a height h. It makes elastic collision with the floor bouncing back to the same height. The frequency of oscillation of its periodic motion is

Moderate

A

$f = 2 \sqrt{\frac{2 h}{g}}$
B

$f = \frac{1}{2 π} \sqrt{\frac{2 h}{g}}$
C

$f = \frac{1}{2} \sqrt{\frac{g}{2 h}}$
D

$f = \frac{1}{2 π} \sqrt{\frac{h}{g}}$

Solution

The body of mass m is at rest on top. It undergoes free fall and attains a speed of v1 at the bottom.
According to the law of conservation of energy, $m g h = \frac{1}{2} m v_{1}^{2}$ .
On simplifying this we get $v_{1} = \sqrt{2 g h}$ …………..(1)
Using this we find the time taken to go down the inclined plane using equations of motion.
$v = u + a t$ , Substituting $v = \sqrt{2 g h}$ , $u = 0$ and $a = g$ ,
We get $\sqrt{2 g h} = u + g t$ ,
Therefore, $\frac{\sqrt{2 g h}}{g} = t$ .
This is time of descent. The total time taken to complete one full oscillation will be two times the time taken to reach the ground.
Hence $\frac{2 \sqrt{2 g h}}{g} = T$
i.e. $T = 2 \sqrt{\frac{2 h}{g}}$ and $f = \frac{1}{2} \sqrt{\frac{g}{2 h}}$

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