A 75 g block is placed into oscillation of amplitude 22 cm as it hangs from a vertical spring.If it observed to have a maximum speed of 14 m/s as it travels through the equilibrium position,what is the spring constant?
A) 300 N/m
B) 22 N/m
C) 4.8 N/m
D) 1200 N/m

Question

Quizplus · Accepted Answer

The correct answer is A because the spring constant is given by k = mω^2, where m is the mass of the block and ω is the angular frequency of the oscillation. The angular frequency is given by ω = 2πf, where f is the frequency of the oscillation. The frequency is given by f = 1/T, where T is the period of the oscillation. The period is given by T = 2π√(m/k). Substituting these equations into the equation for the spring constant, we get k = 4π^2m/T^2. Substituting the given values into this equation, we get k = 4π^2(0.075 kg)/(0.22 m)^2 = 300 N/m.

A 75 G Block Is Placed into Oscillation of Amplitude