Question

A horizontal spring system has a spring with constant 611 N/m and a block of mass 7.97 kg, lying on a frictionless surface, attached to the free end. The a distance from equilibrium and released . Calculate the following 1) angular frequency. 2)the frequency. 3) the period of the oscillating spring system

Answer 1

The angular frequency of the spring system is approximately 8.090 rad/s. The frequency is approximately 1.289 Hz. The period of the oscillating spring system is approximately 0.775 s.

Step-by-step explanation:

The angular frequency (ω) of an oscillating spring system can be calculated using the formula:

ω = √(k/m)

where k is the spring constant and m is the mass of the block.

Plugging in the values given:

k = 611 N/m and m = 7.97 kg

ω = √(611/7.97)

ω ≈ 8.090 rad/s

To find the frequency (f), we use the formula:

f = ω/(2π)

Plugging in the value of ω:

f ≈ 8.090/(2π) ≈ 1.289 Hz

The period (T) of the oscillating spring system can be found using the formula:

T = 1/f

Plugging in the value of f:

T ≈ 1/1.289 ≈ 0.775 s