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

`v(0)=2m/s`

Step-by-step 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*(1m)/(100cm) =0.1m`

The angular frequency can be found using the next equation:

`\omega=\sqrt{(k)/(m) }`

Using the data provided:

`\omega=\sqrt{(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`