A 0.540-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.540-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.2 cm. (Assume the position of the object is at the origin at t = 0.)

(a) Calculate the maximum value of its speed. 43.12 Correct: Your answer is correct. cm/s

(b) Calculate the maximum value of its acceleration. cm/s2

(c) Calculate the value of its speed when the object is 9.20 cm from the equilibrium position. 24.59 Correct: Your answer is correct. cm/s

(d) Calculate the value of its acceleration when the object is 9.20 cm from the equilibrium position.

(e) Calculate the time interval required for the object to move from x = 0 to x = 5.20 cm. s

Explanation / Answer

equation of SHM x = A*sin(wt)

speed v = dx/dt = A*w*cos(wt) = A*w*sqrt(1-(coswt)^2 )

v = A*w*sqrt(1-x^2/A^2)

v = w*sqrt(A^2-x^2)

at x = 0 speed is maximum

Vmax = A*w

vmax = A*sqrt(k/m)

Vmax = 0.112*sqrt(8/0.54) = 43.12 cm/s

_

acceleration a= dx/dt

a = dv/dt = -w^2*sinwt = -w^2*x

amax = -w^2*A = (k/m)*A = = 165.93 cm/s^2

(c)

v = w*sqrt(A^2-x^2)

v = sqrt(8/0.54)*sqrt(0.112^2-0.092^2) = 24.59 cm/s

(d)

a = w^2*x = (8/0.54)*0.092 = 136.3 cm/s^2

(e)

x = A*sin(wt)

w = sqrt(8/0.54) = 3.85

5.2 = 11.2*sin(3.85t)

t = 0.125 s <<<<-----answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.