Question

4) The maximum velocity of a particle executing SHM is 12.56cm/s .Find the amplitude of oscillation if the period of oscillation is 8s .​

Answer 1

Answer:

https://www.toppr.com/ask/question/the-maximum-velocity-of-a-particle-performing-shm-is-628-cms-if-thelength-of-its/

The maximum velocity of a particle performing S.H.M. is 6.28 ...

Answer 2

Step-by-step Explanation:

Given: maximum velocity of the particle ( v ) = 12.56cm/s

period of oscillation ( T ) = 8s

To Find: amplitude of the oscillation

Solution:

• Calculating amplitude of the oscillation

The velocity of a particle executing Simple Harmonic Motion (SHM) is given by;

$v = \omega \sqrt{a^{2} - y^{2}}$  

where a is the amplitude and y is the displacement of the oscillation at the time ( t )

At the mean position, the displacement y = 0 and the velocity will be $v = \omega a$  (maximum velocity).

Therefore, at the maximum velocity of 12.56cm/s and time period of 8s, the amplitude of the oscillation will be;

$a = \frac{v}{\omega} = \frac{T v}{2 \pi}$  

$\Rightarrow a = \frac{8 \times 12.56}{2 \pi} = 16 \ cm$

Hence, the amplitude of oscillation is 16 cm