**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.