Question

A solid copper cube is attached to spring-like device and is able to oscillate horizontally with negligible friction. Each side of the cube is 1.50 cm long. The cube is initially pulled to a point where it stretches the spring-like device by 2.65 cm where it is held at rest with a horizontal force of 1.50 N. What is the frequency at which it oscillates when it is released? (Assume the density of copper is 8.92 g/cm3.)

Answer 1

Answer:

6.91 Hz

Explanation:

Volume of the cube

= (1.5 x 10⁻² )³m³

= 3.375 x 10⁻⁶ m³

mass of the cube

= 3.375 x 10⁻⁶  x 8920 ( 8.92 g / cm³ = 8920 kg/m³ )

m = 30.105 x 10⁻³ kg

Spring stretches by 2.65 x 10⁻² m due to a force of  1.5 N.

If k be the force constant

k x = F

K x2.65 x 10⁻²  = 1.5

k = .566 x 10² N / m

Now frequency of oscillation for spring - mass system is given by

n = $\frac{1}{2\pi} \sqrt{\frac{k}{m} }$

=$\frac{1}{2\pi} \sqrt{\frac{56.6}{.03} }$

n = 6.91 Hz