Question

The maximum velocity attained by the mass of a simple harmonic oscillator is 10 cm/s, and the period of oscillation is 2 s. If the mass is released with an initial displacement of 2 cm, find (a) the amplitude, (b) the initial velocity, (c) the maximum acceleration, and (d) the phase angle.

Answer 1

Answer:Amplitude a = 2 cm, Initial velocity u = 0, $acceleration = -2\pi^{2}$ and $(\pi\times 5+\phi) = \dfrac{\pi}{2}$,  $\phi = \dfrac{-9\pi}{2}$

Explanation:

Given that, velocity v = 10 cm/s

Period of oscillation T = 2 s

Initial displacement y = 2 cm

We know that,

(a). The initial displacement is equal to the amplitude.

Amplitude a = 2 cm

(b). The initial velocity is zero.

$u = 0$

(c). The maximum acceleration is

Acceleration = $(2\pi f)^{2}y$

$acceleration = -2\pi^{2}$

(d). The phase angle is

We know that,

$y = a sin(\omega \ t+\phi)$

Where, $\phi$ = phase angle

$2 =2 sin(\omega \ t+\phi)$

$sin\dfrac{\pi}{2} = sin (\omega t+\phi)$

$(\omega t+\phi) = \dfrac{\pi}{2}$

$(\pi\times 5+\phi) = \dfrac{\pi}{2}$

$\phi = \dfrac{-9\pi}{2}$

Hence, this is the required solution.