An ideal spring is suspended from the ceiling. A 1.80 kg mass is
then attached to the spring causing it to stretch 9.80 cm. The mass is
pulled down 8.00 cm below the equilibrium position and released from rest.
The wave function describing the motion is
y(t)=Asin(wt+fi)
The clock (t=0) is not turned on until the mass reaches y=+4.00 cm.
Take the positive y-direction as up.
(a) [5 pts] Calculate the force constant, k.
kDy
=
mg
k
=
(1.80 kg)(9.80 N/kg)
0.0980 m
=
180. N/m
(b) [5 pts] Calculate the angular velocity, w, of the
oscillation.
k
=
mw2
w
=
é ë
180. N/m
1.80 kg
ù û
1/2
=
10.0 rad/s
(c) [5 pts] Calculate the period, T, of the oscillation.
T
=
2p
w
=
2p rad
10.0 rad/s
=
0.628 s
(d) [5 pts] Calculate the phase constant, fi.
At t=0, y=+4.00 cm.
y
=
Asin(wt+fi)
+4.00 cm
=
(8.00 cm)sin(0+fi)
f
=
p/6 rad
º
30.0o
(e) [5 pts] Determine the speed of the mass when it first passes
through y=+4.00 cm and STATE the direction of the acceleration at that
position.
v
=
w
Ö
A2-y2
=
(10.0 rad/s)[(8.00 cm)2-(4.00 cm)2]1/2
=
69.3 cm/s
Also,
v
=
dy
dt
=
Awcos(wt+p/6)
v(t=0)
=
(8.00 cm)(10.0 rad/s)cos[0+p/6]
=
69.3 cm/s
The figure shows two unattached masses setting on a horizontal air
track. The lower mass is attached to an ideal spring.
k=975 N/m
mB=5.00 kg
mC=10.0 kg
U=7.02 J
An external force is applied to the system stretching the spring. The external
force is removed and the system is released. The amount of energy put into the
system by the external force is U. Assume that the upper block does not move
when the system is in motion (coefficient of static friction ms
between the two blocks).
(a) [6 pts] Calculate the amplitude of the subsequent oscillation.
U
=
1
2
kA2
A
=
æ Ö
2U
k
=
é ë
2(7.02 J)
975 N/m
ù û
1/2
=
0.120 m
(b) [6 pts] Calculate the period of the oscillation.
w
=
Ö
k/mtot
=
æ Ö
975 N/m
15.0 kg
=
8.06 rad/s
T
=
2p
w
=
2p rad
8.062 rad/s
=
0.779 s
Also, T=2pÖ{m/k}.
(c) [6 pt] Calculate the maximum speed of the system.
vmax
=
Aw
=
(0.120 m)(8.06 rad/s)
=
0.967 m/s
(d) [7 pts] Calculate the minimum value the coefficient of static
friction must have so that block C does not slip.
amax
=
Aw2
=
(0.120 m)(8.06 rad/s)2
=
7.80 m/s2
fs(max)=msmCg=mCamax
ms(min)
=
amax
g
=
7.80 m/s2
9.80 m/s2
=
0.796
The wavefunction for a traveling wave on a thin rope is
y(x,t)=(2.30 mm)cos[(6.98 rad/m)x+(742 rad/s)t]
You measure the rope to have a length of 1.35 m and a mass of 0.00338 kg.
Determine the following:
(a) [5 pts] the signal amplitude;
A=2.30 mm
(b) [5 pts] the signal frequency;
f=w/2p = (742 rad/s)/2p = 118 Hz
(c) [5 pts] the signal wavelength;
l = 2p/k=2p/(6.98 rad/m)=0.900 m
(d) [5 pts] the tension in the rope;
m = m/L=(0.00338 kg)/(1.35 m)=0.00250 kg/m
T
=
mc2
=
0.00250 kg/m)[(742 rad/s)/(6.98 rad/s)]2
=
28.3 N
(e) [5 pts] the average power transmitted by the wave.
One string of a musical instrument is bounded between two supports
75.0 cm apart. This section has a mass of 8.75 g.The string is played in
the third harmonic (second overtone). An observer in the room hears a sound
having a wavelength lair=0.765 m (cair=344 m/s).
(a) [5 pts] Calculate the mass density of the string.
m
=
m
l
=
8.75×10-3 kg
0.750 m
=
1.17×10-2 kg/m
(b) [5 pts] Calculate the frequency of the signal generated in the
string.
The frequency generated in the string is the same as the frequency generated
in air.
fair
=
cair
lair
=
344 m/s
0.765 m
=
4.50×102 Hz
=
fstring
(c) [10 pts] Determine the wavelength of the signal in the string and
compute the wave speed in the string.
The wavelength of the signal in the string is
lstring
=
2L/n
=
2(0.750 m)
3
=
0.500 m
cstring
=
l3f3
=
(0.500 m)(450 Hz)
=
225 m/s
(d) [5 pts] What is the frequency of the first harmonic string signal?
f3
=
3f1
f1
=
(4.50×102 Hz)/3
=
150 Hz
FORMULAS
E=kA2/2
v=w/k
t = rFsinq
k=2p/l
å
Fx=0
å
Fy=0
å
tz=0
fn=nf1
x(t)=Asin(wt+f)
w = 2pf
T=2p/w
F=-kx
K=mv2/2
m = m/L
vmax=Aw
amax=Aw2
U=kx2/2
E=K+U
P=
Ö
mF
w2A2/2
v=lnfn
y(x,t)=Asin(kx-±wt+f)
ln=2L/n
W=mg
w =
Ö
k/m
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version 3.85. On 19 Jan 2016, 10:59.