Answer to Question #162442 in Physics for nasia

Question

A particle execute simple harmonic motion about the point x = 0. At t= 0, it has

displaced y = 0.37 cm, and zero velocity. If the frequency of motion is 0.25/s. Calculate

the period, amplitude, the maximum speed and the maximum acceleration.

Answer 1

The equation of SHM in general form is:

$$y(t) = A\cos(\omega t)$$

where $A = 0.37cm$ is the amplitude, $\omega = 2\pi f = 2\pi\cdot0.25rad/s = 0.5\pi \space rad/s$ is the angular frequncy. The period is:

$$\omega T = 2\pi\\ T = \dfrac{2\pi}{\omega} = 4s$$

The speed is (and it is 0 at t=0):

$$\dot{y}(t) = -A\omega\sin(\omega t)$$

And the maximum speed is:

$$\dot{y}_{max} = A\omega = 0.175\pi \space cm/s$$

The acceleration is:

$$\ddot{y}(t) = -A\omega^2\cos(\omega t)$$

And the maximum acceleration is:

$$\ddot{y}_{max} = A\omega^2 = 0.37\cdot (0.5\pi)^2 =0.0925\pi^2 \space cm/s^2$$

Answer. Period: $4s$ , amplitude: $0.37cm$, maximum speed $0.175\pi \space cm/s$, maximum acceleration: $0.0925\pi^2 \space cm/s^2$