A mass m = 0.75 kg hangs at the end of a vertical spring whose top end is fixed

Question

A mass m = 0.75 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 105 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(omega t - phi). The positive y-axis points upward. Find the angular frequency of oscillation, in radians per second. Determine the value of the coefficient A, in meters. Determine the value of phi, in radians. Enter an expression for the velocity along the y-axis as a function of time, in terms of A, omega, and t, using the value for phi from the previous part. What is the mass's velocity, in meters per second, at time t_1 = 0.25 s? What is the magnitude of the mass's maximum acceleration, in meters per second squared?

Explanation / Answer

B) A = 0.35m

D) Vy = - Aw sin (wt - phi)

put the value of phi found in part c

E) Vy = - 0.35*w sin (0.25w - phi)

put the value of w from a and phi from c and calculate velocity.

F) a max = Aw^2 = Ak/m = 0.35* 105/0.75

= 49 m/s^2

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