Question

A block of 5.0kg is attached to end of a spring constant 230N/m. The spring mass-system undergoes a simple harmonic motion. At t=0, the position of the block is x=0.23m,and its velocity is v = -5.4 m/s.

What is the angular frequency?

What is the position x as a function of time.(form of x =A sin(wt +0)

Answer 1

The angular frequency is

$$\omega=2\pi f=2\pi=\sqrt{k/m}=6.78\text{ rad/s}.$$

Applying energy conservation, find the amplitude:

$$\frac12kx^2+\frac12mv^2=\frac12mA^2,\\\space\\ A=\sqrt{v^2+\frac{kx^2}m}=5.62\text{ m}.$$

The position as a function of time is

$$x =A \sin(\omega t +\phi),\\ 0.23=5.62\sin(6.78·0+\phi),\\ \phi=2.24°\equiv0.0391\text{ rad}.\\\space\\ x=5.62\sin(6.78t+0.0391).$$