Question

An object of mass 7.4 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field. The spring stretches 7.8 cm before it reaches its equilibrium position. If this system is allowed to oscillate, what will be its frequency? (Give your answer to the nearest 0.1 Hz)

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the object = m = 7.4 kg

Spring constant = k

Extension of the spring due to the object = X = 7.8 cm = 0.078 m

kX = mg

k(0.078) = (7.4)(9.81)

k = 930.69 N/m

Frequency of the oscillation = f

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

$f = \frac{1}{2\pi }\sqrt{\frac{930.69}{7.4}}$

f = 1.8 Hz

Frequency of oscillation of the system = 1.8 Hz