Question

A block of mass mmm= 3.00 kg is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance hhh= 8.00 cm from its equilibrium length. (Figure 1)The spring has an unknown spring constant k. Take the acceleration due to gravity to be g = 9.81 m/s2m/s2 .

Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting frequency ff of the block's oscillations about its equilibrium position.

Express your answer in hertz.

Answer 1

Solution) m = 3 kg

h = 8 cm = 8×10^(-2) m

g = 9.81 m/s^2

f = ?

Force , F = KX

Here K is spring constant

F = mg

X = h

=> mg = Kh

3×9.81 = (K)(8×10^(-2))

3×9.81 = (K)(0.08)

K = (3×9.81)/(0.08)

K = 367.875 N/m

K = 367.9 N/m

Frequency , f = (1/(2(pi)))(K/m)^(1/2)

f = (1/(2(pi)))(367.9/3)^(1/2)

f = 1.76 Hz