April 18, 2008
PHY2053 Discussion
Quiz 11 (Chapter 13.5-13.11)
Name:
UFID:
**1. (5 pts) A light balloon filled with helium of density 0.180 kg/m³ is tied to a light
string with a length of 1.50 m. The string is fixed to the ground, forming an inverted
simple pendulum. If the balloon is displaced slightly from equilibrium, the motion is
simple harmonic. Find the period of the motion. (Density of the air is 1.29 kg/m³.)
The component of the net force along the direction of motion is
F = -(B-mg)sinθ ≈ -(ρVg-ρ’Vg)s/L
Since the force is proportional to the displacement, the motion is simple harmonic. The
acceleration of the balloon is given by
a = F/m = F/(ρ’V) = -(ρg-ρ’g)/(ρ’L)×s.
Using a = -ω²s , we have
ω = √(((ρ-ρ’)g)/(ρ’L))
The period is related to the angular frequency:
T = 2π/ω = 2π√((ρ’L)/ (ρ-ρ’)g) = 2×3.14×√((0.18×1.5)/(9.8×(1.29-0.18))) = 0.989 s
*2. (5 pts) Ocean waves are traveling eastward with a frequency of 0.500 Hz. The waves
have a wavelength of 20.0 m. What is the apparent frequency observed by a boat
traveling eastward at 3.00 m/s?
The speed of the wave is given by
v = fλ = 0.5×20 = 10 m/s
Since the observed frequency is the number of waves that pass the observer in one
second, we need the relative velocity of the wave relative to the boat:
V = v-v’ = 10-3 = 7 m/s
The observed frequency is
f = V/λ = 7/20 = 0.350 Hz.
**3. (5 pts) A string with a mass of 15.0 g and a length of 2.50 m has its end tied to two
walls at the same height. The walls are separated by a distance of 2.00 m. Two objects,
each with a mass of 3.00 kg, are suspended from the string at two points 0.500 m away
from its ends. Find the speed of the transverse wave in the horizontal segment of the
string.
First we calculate the angle the inclined section of the string makes to the horizontal.
Since the horizontal section of the string is 1.5 m, the angle is
θ = cosˉ¹(0.25/0.5) = 60˚
The tension in the horizontal segment is given by the equilibrium condition at an end of
the segment:
y: T’sin60˚-mg = 0 ⇒ T’ = mg/sin60˚
x: T-T’cos60˚ = 0 ⇒ T = (mg/sin60˚)cos60˚ = mg/tan60˚ = 3×9.8/tan60˚ = 17.0 N
The linear density is given by
μ = m/L = 15×10ˉ³/2.5 = 6×10ˉ³ kg/m
Thus the speed of the wave is
v = √(T/μ) = √(17/(6×10ˉ³)) = 53.2 m/s
*4. (5 pts) A wave of amplitude 0.20 m interferes with a second wave of amplitude 0.15
m traveling in the same direction. What is the smallest resultant amplitude that can
occur?
The minimum amplitude occurs when the two waves interfere destructively, and the
amplitude is
A = A₁-A₂ = 0.2-0.15 = 0.05 m.