Question

A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harmonic motion with an amplitude of 10 cm. The speed of the 1.0 kg mass at the equilibrium position is?

Answer 1

Answer:

$v(0)=2m/s$

Explanation:

The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

$v(t)=\omega *A*cos(\omega t + \phi)$

Where:

$\omega=Angular\hspace{3}frequency\\A=Amplitude\\\phi=Initial\hspace{3}phase$

Express the amplitude in meters:

$10cm*\frac{1m}{100cm} =0.1m$

The angular frequency can be found using the next equation:

$\omega=\sqrt{\frac{k}{m} }$

Using the data provided:

$\omega=\sqrt{\frac{400}{1} } =20$

At the equilibrium position:

$\phi=0$

$v(0)=20*(0.1)cos(20*0+0)=2*cos(0)=2*1=2m/s$