A mass m 0.55 kg hangs at the end of a vertical spring whose top end is fixed to

Question

A mass m 0.55 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 75 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y(t) = A cos(omega t - phi), with the positive y axis pointing upward. At time t = 0 the mass is observed to be at a distance d = 0.45 m below its equilibrium height with an upward speed of v_o = 4 m/s. Find the angular frequency of the oscillation, in radians per second. Find the value of p, in radians Calculate the value of A, in meters. What is the mass's velocity along the y axis, n meters per second, at time t 0.25a? What is the magnitude of the mass's maximum acceleration, in meters per second squared?

Explanation / Answer

y = A * cos(w * t - phi)

part a)

w = sqrt(k/m)

w = sqrt(75/.55)

w = 11.7 rad/s

part B)

let the amplitude is A

0.5 * 75 * A^2 = 0.5 * 75 * 0.45^2 + 0.5 * 0.55 * 4^2

solving for A

A = 0.57 m

-0.45 = 0.57 * cos(11.7 * 0 - phi)

phi = 2.48 radian

the angle phi is 2.48 radian

part D)

at t1 = 0.25 s

V = 0.57 * 11.7 * cos(0.25 * 11.7 + 2.48)

V = 4.25 m/s

part E)

for the maximum acceleraration

maximum acceleration = 11.7^2 * 0.57

maximum acceleration = 78.03 m/s^2

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