Sample Problem

Sample Problem

This maximum speed occurs when the oscillating block is rushing
through the origin; compare Figs. 16-4a and 16-4b, where you can
see that the speed is a maximum whenever x = O.
Sample Problem 16-1
A block whose mass m is 680 g is fastened to a spring whose spring
constant k is 65 N/m. The block is pulled a distance x = II cm
from its equilibrium position at x = 0 on a frictionless surface and
released from rest at t = O.
(a) What are the angular frequency, the frequency, and the period
of the resulting motion?
SOLUTION: The Key Idea here is that the block-spring system forms
a linear simple harmonic oscillator, with the block undergoing
SHM. Then the angular frequency is given by Eq. 16-12:
W = Vf;£;; =.)65 N/m = 9.i8 rad/s 0.68 kg
= 9.8 rad/s. (Answer)
The frequency follows from Eq. 16-5, which yields
W 9.78 rad/s
f = 21T = 21T rad = 1.56 Hz = 1.6 Hz. (Answer)
The period follows from Eq. 16-2, which yields
I I
T = – = — = 0.64 s = 640 ms. f 1.56 Hz
(Answer)
(b) What is the amplitude of the oscillation?
SOLUTION: The Key Idea here is that, with no friction involved, the
mechanical energy of the spring-block system is conserved. The
block is released from rest II cm from its equilibrium position,
with zero kinetic energy and the elastic potential energy of the
system at a maximum. Thus, the block will have zero kinetic energy
whenever it is again II ern from its equilibrium position, which
means it will never be farther than 11 ern from that position. Its
maximum displacement is 11 ern:
xm = II cm. (Answer)
(c) What is the maximum speed Vm of the oscillating block, and
where is the block when it occurs?
SOLUTION: The Key Idea here is that the maximum speed Vm is the
velocity amplitude WXm in Eq. 16-6; that is,
Vm = WXm = (9.78 rad/s)(O.11 m)
= J.I m/s. (Answer)
(d) What is the magnitude am of the maximum acceleration of the
block?
SOLUTION: The Key Idea this time is that the magnitude am of the
maximum acceleration is the acceleration amplitude wlx., in
Eq. 16-7; that is,
am = w1x •. = (9.78 rad/s)2(0.11 m)
= II m/s2. (Answer)
This maximum acceleration occurs when the block is at the ends
of its path. At those points, the force acting on the block has its
maximum magnitude; compare Figs. 16-40 and 16-4c, where you
can see that the magnitudes of the displacement and acceleration
are maximum at the same times.
(e) ~at is the phase constant 41for the motion?
SOLUTION: Here the Key Idea is that Eq. 16-3 gives the displacement
of the block as a function of time. We know that at time t = 0, the
block is located at x = xm• Substituting these initial conditions, as
they are called, into Eq. 16-3 and canceling Xm give us
I = cos 41.
Taking the inverse cosine then yields
41 = 0 rad.
(16-14)
(Answer)
(Any angle that is an integer multiple of 21T rad also satisfies
Eq. 16-14; we chose the smallest angle.)
(f) What is the displacement function x(t) for the springblock
system?
SOLUTION: The Key Idea here is that x(t) is given in general form by
Eq. 16-3. Substituting known quantities into that equation gives us
x(t) = x”, cos(wt + 41)
= (0.11 m) cos[(9.8 rad/s)t + 0]
= 0.11 cos(9.8t), (Answer)
where x is in meters and t is in seconds.

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