Question

An object moves with simple harmonic motion. If the amplitude and the period are both doubled, the object's maximum speed is 1. quartered 2. halved 3. quadrupled. 4. doubled 5. unchanged

Answer 1

Answer:

The object's maximum speed remains unchanged.

Explanation:

The speed of a particle in SHM is given by :

$v(t)=A\omega\ sin(\omega t)$

Maximum speed is, $v_{max}=A\omega$

If A' = 2A and T' = 2T

$v'_{max}=(2A)\dfrac{2\pi}{2T}$

$v'_{max}=(A)\dfrac{2\pi}{T}$

$v'_{max}=v_{max}$

So, the maximum speed of the object remains the same i.e. it remains unchanged. Hence, this is the required solution.