Question

Two mass executing shm under the influence of vertical spring as shown with amplitude of 2 cm and angular frequency of 5 radius per second in the extension of the spring is maximum suddenly it ends breaks and 3kg mass falls off what is the maximum velocity of 1 kg mass

Answer 1

Given:

Two masses are executing SHM under the influence of a vertical spring with an amplitude of 2 cm and angular frequency of 5 rad/s.

At max extension , the 3kg block breaks and falls off.

To find:

Maximum velocity of 1kg Mass.

Calculation:

We need to apply the concept of Conservation of Mechanical Energy in order to solve this problem.

The maximum potential spring energy at the amplitude will be converted into maximum kinetic energy for the 1 kg block.

$\therefore \: PE = KE$

$= > \: \dfrac{1}{2} m { \omega}^{2} {a}^{2} = \dfrac{1}{2} (m2) {v}^{2}$

$= > \: \cancel{ \dfrac{1}{2}} m { \omega}^{2} {a}^{2} = \cancel{ \dfrac{1}{2} }(m2) {v}^{2}$

$= > \: m { \omega}^{2} {a}^{2} = (m2) {v}^{2}$

$= > \: (3 + 1) \times {(5)}^{2} \times {(0.02)}^{2} = (1) \times {v}^{2}$

$= > \: 4\times 25 \times {(0.02)}^{2} = {v}^{2}$

$= > {v}^{2} = \sqrt{ {2}^{2} \times {5}^{2} \times {(0.02)}^{2} }$

$= > \: v = 2 \times 5 \times 0.02$

$= > \: v = 0.2 \: m {s}^{ - 1}$

So , final answer is :

Maximum velocity of 1 kg block is 0.2 m/s.