contestada

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 0.120 m. The mass is now displaced from its equilibrium position and undergoes simple harmonic oscillations. How long does it take the mass to complete one full oscillation?

Respuesta :

Answer:

Time period of the oscillation will be 0.695 sec

Explanation:

We have given mass attached to the string m = 0.350 kg

The spring is stretched by 0.120 m

So x = 0.120 m

We know that force is given by F = Kx

So mg = kx

[tex]0.350\times 9.8=k\times 0.120[/tex]

k = 28.5833

Now time period is given by

[tex]T=2\pi \sqrt{\frac{m}{k}}=2\times 3.14\sqrt{\frac{0.35}{28.5833}}=0.695sec[/tex]

So time period of the oscillation will be 0.695 sec

Otras preguntas