A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 115 N/m and negligible mass. At time t = 0 the mass is released from rest at a distance d = 0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by y(t) = A cos(ωt – φ). The positive y-axis points upward. show answer Correct Answer 17% Part (a) Find the angular frequency of oscillation, in radians per second. ω = 10.22 ✔ Correct! show answer Incorrect Answer 17% Part (b) Determine the value of the coefficient A, in meters.

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barackodam barackodam · Answer 1 · 2020-04-23T10:57:29+02:00

Answer:

a) = 10.22 rad/s

b) = 0.35 m

Explanation:

Given

Mass of the particle, m = 1.1 kg

Force constant of the spring, k = 115 N/m

Distance at which the mass is released, d = 0.35 m

According to the differential equation of s Simple Harmonic Motion,

ω² = k / m, where

ω = angular frequency in rad/s

k = force constant in N/m

m = mass in kg

So,

ω² = 115 / 1.1

ω² = 104.55

ω = √104.55

ω = 10.22 rad/s

If y(0) = -0.35 m and we want our A to be positive, then suffice to say,

The value of coefficient A in meters is 0.35 m