A particle with a mass of 0.420 kg is attached to a horizontal spring with a for

Question

A particle with a mass of 0.420 kg is attached to a horizontal spring with a force constant of 34.02 N/m. At the moment t = 0, the particle has its maximum speed of 22.5 m/s and is moving to the left. (Assume that the positive direction is to the right.)

(a) Determine the particle's equation of motion, specifying its position as a function of time. (Use the following as necessary: t.)

(b) Where in the motion is the potential energy three times the kinetic energy?


(c) Find the minimum time interval required for the particle to move from x = 0 to x = 1.00 m. (Be sure to enter the minimum time and not the total time elapsed.)


(d) Find the length of a simple pendulum with the same period.

Explanation / Answer

a)

x = a*sin(wt)

w = sqrt (k/m) = sqrt (34.02 / 0.42) = 9 rad/s

Max speed v = a*w

-22.5 = a*9

a = -2.5 m

So, x = -2.5*sin(9t)

b)

v = dx/dt = -2.5*9*cos(9t) = -22.5*cos(9t)

KE at time t = 1/2*mv^2

PE at time t = 1/2*kx^2

1/2*kx^2 = 3*1/2*mv^2

v^2 = (1/3)*(k/m)*x^2

But k/m = w^2

So, v = wx / (sqrt3)

-22.5*cos(9t) = 9*(-2.5*sin(9t))

Tan(9t) = 1

t = 0.08727 s

c)

Time from x = 0 to 1 will be equal to time from 0 to -1.

At x = -1, we have -1 = -2.5*sin(9t)

t = 0.0457 s

d)

w = sqrt (g/L) = 9

sqrt (9.81 / L) = 9

L = 0.1211 m

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