A solid copper cube is attached to spring-like device and is able to oscillate horizontally with negligible friction. Each side
1 answer:
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 =
=
n = 6.91 Hz
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