(dl9564) lew hk9 Opyrchal (41104) This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation A = A0 sin t + 3 ,
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When the mass moves out of equilibrium, it suers a net restoring force
net Fy = F spring M g = k(y y eq ) = ky ,
and accelerates back towards the equilibrium position at the rate
net Fy k ay = = y. M M
where = radians per second, t is in seconds, and A0 = 6 m. What is the phase of the motion at t = 8.8 s? Correct answer: 28.6932 rad. Explanation: Let : t = 8.8 s = . and
Therefore, the mass oscillates harmonically with angular frequency = ay = y k = M 1 105 N/m 15.3 kg
= 80.8452 s1 .
x = A0 sin( t + ) The phase is the angle in the argument of the sine function, and from the problem statement we see it is = t+ 3 = ( rad/s) (8.8 s) + 3 = 28.6932 rad . 002 (part 1 of 4) 10.0 points A 15.3 kg mass is suspended on a 1 105 N/m spring. The mass oscillates up and down from the equilibrium position yeq = 0 according to y(t) = A sin(t + 0 ) . Find the angular frequency of the oscillating mass. Correct answer: 80.8452 s1 . Explanation: Let : M = 15.3 kg and k = 1 105 N/m .
003 (part 2 of 4) 10.0 points At time t0 = 0 the mass is at 6.8 cm and moving upward at velocity +17.2 m/s. Find the amplitude of the oscillating mass. Correct answer: 22.3355 cm. Explanation:
Let : y0 = 6.8 cm and v0 = 17.2 m/s . The mass oscillates according to the SHM equation y(t) = A sin(t + 0 ) , so its velocity is vy (t) = dy = A cos(t + 0 ) . dt
At time t0 = 0, we have y0 y(t = 0) = A sin 0 and so v0 . (1)
v0 = vy (t = 0) = A cos 0 , A sin 0 = y0 and A cos 0 =
lew (dl9564) hk9 Opyrchal (41104) Consequently, A2 = (A sin 0 )2 + (A cos 0 )2 v0 2 = (y0 )2 + v0 2 A = (y0 )2 + = (6.8 cm)2 + 17.2 m/s 80.8452 s1
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A large block with mass 15 kg executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency 2.03 Hz . A smaller block with mass 5 kg rests on it, as shown in the gure, and the coecient of static friction between the two is s = 0.562 . 5 kg k s = 0.562 15 kg
= 22.3355 cm . 004 (part 3 of 4) 10.0 points Find the initial phase 0 in the range 0 0 < 360 . Correct answer: 17.725. Explanation: According to eq. (1), A sin 0 y0 tan 0 = = = 0.319621 . A cos 0 v0 y0 > 0 and v0 > 0 imply sin 0 > 0 , cos 0 > 0 , and 0 < 0 < 90 . Consequently, the initial phase is 0 = arctan(0.319621) = 17.725 . 005 (part 4 of 4) 10.0 points Calculate the position of the oscillating mass at the time t = 0.0723 s . Correct answer: 2.86705 cm. Explanation: Let : t = 0.0723 s . where
What maximum amplitude of oscillation can the system have if the top block is not to slip? The acceleration of gravity is 9.8 m/s2 . Correct answer: 3.3854 cm. Explanation: Let : mb mt f s g = 15 kg , = 5 kg , = 2.03 Hz , = 0.562 , and = 9.8 m/s2 . N fs mg For harmonic simple motion, we have x = A cos( t + ) and
a = A 2 cos( t + ) , y = A sin(t + 0 ) = (22.3355 cm) sin[(80.8452 s1 )(0.0723 s) + 17.725 ] = 2.86705 cm . 10.0 points
= 2 f = 2 (2.03 Hz) = 12.7549 rad/s and since the extreme values of the cosine are 1, amax = A 2 . The force of friction is fs N .
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lew (dl9564) hk9 Opyrchal (41104) Horizontally, the large block only feels the force of friction, so fs = m a and the maximum frictional force is fs = s N = s m g since the setup is horizontal. (If it were not, the normal force would also depend on the angle of inclination.) From Newtons 2nd law F = ma s m g = m Amax 2 , so s g Amax = 2 (0.562)(9.8 m/s2 ) 100 cm = (12.7549 rad/s)2 1m = 3.3854 cm . Note: Neither mass is required to work this problem. keywords: 007 10.0 points The spring constant of the spring in the gure is 24 N/m, and the mass of the object is 8 kg. The spring is unstretched and the surface is frictionless. A constant 28 N force is applied horizontally to the object (stretching the spring). 28 N 24 N/m v0 = 0 8 kg Find the speed of the object after it has moved a distance of 0.15 m. Correct answer: 0.991211 m/s. Explanation: Let : m = 8 kg , F = 28 N , s = 0.15 m , and k = 24 N/m .
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In the presence of non-conservative forces, Wnc = 1 1 2 2 m vf m vi + m g yf 2 2 1 1 m g yi + k x2 k x2 i f 2 2 1 1 2 F s = m vf + k s2 2 2 2 s F k s2 vf = . m
Since 2 s F k s2 = 2 (0.15 m) (28 N) (24 N/m) (0.15 m)2 = 7.86 m N , then vf = 7.86 m N = 0.991211 m/s . 8 kg
008 10.0 points A uniform disk of radius 1.5 m and mass 3.8 kg is suspended from a pivot 0.69 m above its center of mass. The acceleration of gravity is 9.8 m/s2 .
axis
Find the angular frequency for small oscillations. Correct answer: 2.05508 rad/s. Explanation: Basic Concepts The physical pendulum: = I = m g d sin d2 dt2 so that the angular frequency for small oscillations (sin ) is = = mgd I .
lew (dl9564) hk9 Opyrchal (41104) Parallel axis theorem I = I0 + m a 2 Solution: We need the moment of inertia of the disk about the pivot point, which we call P. The moment of inertia of a uniform disk about its center is Idisk = 1 m R2 , 2
ar
4
r
g
a
m v = 0 at
but here the disk is rotating about P, a distance d from the center of mass. The parallel axis theorem lets us move the axis of rotation a distance d: 1 IP = m R 2 + m d2 = m 2 R2 + d2 2 .
If the moons gravity is one-sixth of Earths gravity, nd the length of the pendulum on the moon. Correct answer: 0.503333 m. Explanation: The period of a pendulum is dened by T = 2 . g
Then using the formula for the small angle oscillation frequency of a physical pendulum (see Basic Concepts above), we obtain mgd = IP mgd R2 + d2 m 2
Since the periods are the same, 2 m = 2 gm m = gm g m = gm g g g
=
or gd R2 2 = + d2
= =
=
1 g 6
(9.8 m/s2 ) (0.69 m) (1.5 m)2 + (0.69 m)2 2 = 2.05508 rad/s .
1 6 1 = (3.02 m) 6 = 0.503333 m .
009 10.0 points A pendulum oscillating on the moon has the same period as a(n) 3.02 m pendulum oscillating on Earth.