Review Packet Answers

1.
a) amplitude = 1.5 inches
b) T = 0.6 s, f = 1/T, so f = 5/3 Hz
c) v = f*lambda = (5/3 Hz)*1.4 m = 2.33 m/s
d) using a constant accel. equation, displacement is 1.33 m.
e) We must use the Doppler effect equation: F_obs = (5/3 Hz) * (2.33 m/s + 2 m/s)/(2.33 m/s) = 3.10 Hz

2.
a) either 1.0 meters or 1.07 meters, depending on whether L1 or L2 was used.
b) lambda = v / f so f = 343 Hz.
c) The frequency of the waves will be the same in the water since it doesn't change when crossing from one medium to another. Using the wave equation, the wavelength is 4.34 meters.
d) L3 = 5*lambda/4 = 1.25 meters.
e) The speed of sound increases when temperature increases. If the temperature is higher, then the speed will be higher, which means the actual frequency will be higher. This means the number calculated in (b) is too low.

1982B2.
a) hook has T2 upwards, T1 and m2g downwards. The load has T1 up, m1g downwards.
b) T1 = 6000 N, T2 = 6600 N.

2000B2.
a) T and f directed up the incline, Fn perpendicular to the incline surface, and m1g directed downwards
b) mu = f/(m1g cos(theta))
c) M = (m1 + m2) - 3f/g
d) a = g sin(theta) - f/m1

1995B4.
a) 400 W
b) 40%
c) 1000 W
d) 600 W

2002B5
a) By the right hand rule, F_mag is directed to the left. This means the electric force must go to the right. Since the proton has a positive charge, the electric field must also go towards the right.
b) qE = qvB, so v = E/B.
c) proton moves in a circular path towards the left.
d) a = eE/m

small #1:
a) lambda = 2L
b) v = 2f0L
c) 2f0
d) f = L/2

Last problem with motor:

a) 3 milliwatts (3 mW)
b) 0.18 J
c) work = 0.118 J, so efficiency = actual output/energy in = 65.6 %
d) 5000 ohm and 1000 ohm resistors connected in series with the motor.