Question

(25%) Problem 3: A mass m = 0.65 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 65 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(ot - *), with the positive y-axis pointing upward. At time t = 0 the mass is observed to be passing through its equilibrium height with an upward speed of vo = 4.8 m/s. Otheexpertta.com 20% Part (a) Find the angular frequency of the oscillation, in radians per second Grade Summary Deductions 0% 100% Potential sin cotan atan cosh tan acos sinh cotanh Submissions 7 8 9 НОМЕ Attempts remaining: 30 (2% per attempt) detailed view E A asin acotan tanh ODegrees 4 5 6 1 2 3 0 END Radians VOBACKSPACE DEL CLEAR I give up! Submit Hint Feedback Hints: 1% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. 20% Part (b) Find the smallest positive value of o, in radians - A 20% Part (c) Calculate the value of A, in meters. 20% Part (d) What is the mass's velocity along the y-axis, in meters per second, at time t1 = 0.25 s? 20% Part (e) What is the magnitude of the mass's maximum acceleration, in meters per second squared? All content 2019 Expert TA, LLC

Answer 1

a) w = sqrt(k/m)

= sqrt(65/0.65)

= 10 rad/s <<<<<<<<<-----------Answer

b) at t = 0 ,

y(t) = A*cos(w*0 - phi)

0 = cos(-phi)

cos(pi/2) = cos(phi)

==> phi = pi/2 rad (or) 1.57 rad <<<<<<<<<-----------Answer

c) we know, vo = A*w

==> A = vo/w

= 4.8/10

= 0.48 m <<<<<<<<<-----------Answer

d)
v(t) = dy/dt

v(t) = -vo*sin(w*t - phi)

at t = 0.25 s,

v = -4.8*sin(10*0.25 - pi/2)

= -3.84 m/s <<<<<<<<<-----------Answer

e) a_max = A*w^2

= 0.48*10^2

= 48 m/s^2 <<<<<<<<<-----------Answer