A mass of 0.24 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.24 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.42 m)cos[(6 rad/s)t]. Determine the following.

(a) amplitude of oscillation for the oscillating mass
(b) force constant for the spring
(c) position of the mass after it has been oscillating for one half a period
(d) position of the mass one-third of a period after it has been released
(e) time it takes the mass to get to the position x = 0.10 m after it has been released

Explanation / Answer

Here ,

x = 0.42 * cos(6 rad/s * t)

as x = A * cos(w * t)

a) comparing to the equation ,

A = 0.42 m

the ampitude of the mass is 0.42 m

B)
as w = 6 rad/s

w = sqrt*(k/m )

6 = sqrt(k/.24)

k = 8.64 N/m

the spring constant is 8.64 N/m

c)

at t = T/2

cos(pi) = -1

x = - 0.42 m

the positiion of mass is -0.42 m
d)

after 1/3 period ,

cos(2pi * 1/3) = -0.5

x = -0.5 * 0.42

x = -0.21 m

tre position of the particle is -0.21 m

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