A 0.8 kg block sliding on a horizontal frictionless surface is attached to a hor

Question

A 0.8 kg block sliding on a horizontal frictionless surface is attached to a horizontal spring with k = 500 N/m. Let x be the displacement of the block from the position at which the spring is unstretched. At t = 0 the block passes through x = 0 with a speed of 5.2 m/s in the positive x direction.
(a) What is the frequency of the block's motion?
Hz

(b) What is the amplitude of the block's motion?
m

(c) The displacement x may be written as a function of time, x(t) = xm cos(?t + phi). What are the values of xm, ?, and phi?
x = m
? = rad/s
phi = rad

Explanation / Answer

(a) T = 2(m/k)1/2 = 2(0.8 kg/500 N/m)1/2 = 0.2513 s

Since f = 1/T, f = 3.979 Hz.

(b) The general formula for SHM is x = A cos (t + ).

v = dx/dt = -A sin (t + ) .      We know v, and we can get from the frequency, which is what we need to solve for A, which is a maximum when the sine is 1.   = 2f = 25 s-1.

A = v/ = (5.2 m/s)/(25 s-1) = 0.208 m

(c) xm is the amplitude = 0.208 m

? is = 25 rad/s

phi = = -/2 . To see this, note that we are given that x = 0 when t = 0.  Since t = 0, cos(t + ) = cos() and this must equal zero. Therefore = /2 or -/2. The negative solution is the correct one since x is increasing with time.

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