Question

A particle execute simple harmonic motion about the point x=0.At t=0,it has displaced y=0.37cm & zero velocity.If the frequency of motion is 0.25 Hz.Calculate the period, amplitude,the maximum speed & the maximum acceleration.

Answer 1

(a)

$$T=\dfrac{1}{f}=\dfrac{1}{0.25\ Hz}=4.0\ s.$$

b)

$$A=0.37\ cm=3.7\cdot10^{-3}\ m.$$

c) Let's first find angular frequency:

$$\omega=2\pi f=2\pi\cdot0.25\ Hz=1.57\ \dfrac{rad}{s}.$$

Then, we can write the displacement of the particle:

$$x(t)=0.0037\cdot cos(1.57t).$$

Let's take the derivative of the displacement with respect to $t$:

$$v(t)=\dfrac{d}{dt}(x(t))=-0.0037\cdot1.57sin(1.57t)=-0.0058sin(1.57t).$$

The maximum speed will be when $sin(1.57t)=1.$ Then, we get:

$$v_{max}=-0.0058\ \dfrac{m}{s}.$$

d) Let's take the derivative of the speed with respect to $t$:

$$a(t)=\dfrac{d}{dt}(v(t))=-0.0058\cdot1.57cos(1.57t)=-0.0091cos(1.57t).$$

The maximum acceleration will be when $cos(1.57t)=1.$ Then, we get:

$$a_{max}=-0.0091\ \dfrac{m}{s^2}.$$