A 0.420-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.420-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.4 cm. (Assume the position of the object is at the origin at t = 0.) (a) Calculate the maximum value of its speed. (b) Calculate the maximum value of its acceleration. (c) Calculate the value of its speed when the object is 7.40 cm from the equilibrium position. (d) Calculate the value of its acceleration when the object is 7.40 cm from the equilibrium position. (e) Calculate the time interval required for the object to move from x = 0 to x = 3.40 cm.

Explanation / Answer

In SHM

x = A*sin(wt)

speed v = dx/dt = A*coswt*w

V = A*w*coswt

v = A*w*sqrt(1-sin(wt)^2)

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

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

maximum speed occurs when x = 0

Vmax = Aw

(a)

vmax = A*w = A*sqrt(k/m)

vmax = 0.114*sqrt(8/0.42)

vmax = 0.498 m/s

(b)

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

amax = w^2*A = (k/m)*A =(8/0.42)*0.114 = 2.17

_____________

(c)

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

v = (8/0.42)*sqrt(0.114^2-0.074^2) = 1.65 m/s

(d)

a = w^2*x = (8/0.42)*0.074 = 1.41 m/s^2

(e)

0.034 = 0.114*sin((8/0.42)*t)

t = 17.35 s <<<-----answer

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