August 6, 2008
PHY2053 Discussion
Quiz 10 (Chapter 13.5-14.11)
Name:
UFID:
**1. (5pts) A simple pendulum with a length of 1.50 m is suspended from the ceiling of a
train. What is the period of simple harmonic motion for the pendulum if the train is
moving up an incline of 15.0˚ with an acceleration of 3.00 m/s²?
The pendulum is said to be in equilibrium if the pendulum is stationary relative to the
train. We calculate the tension of the pendulum in equilibrium, and equate it with mgeff.
Taking +x axis along the incline and up and +y perpendicular to the incline, we get
x: ma = Fsinθ-mgsin15˚ ⇒ F²sin²θ = m²(a+gsin15˚)²
y: 0 = Fcosθ-mgcos15˚ ⇒ F²cos²θ = m²cos²15˚
⇒ F = m√((a+gsin15˚)²+ cos²15˚) = mgeff,
where geff = √((a+gsin15˚)²+ cos²15˚) = √{(3+9.8sin15˚)²+cos²15˚} = 11.0 m/s
Thus the period of the pendulum is
T = 2π√(L/geff) = 2×3.14×√(1.5/11) = 2.32 s
*2. (5pts) A stone is thrown downward into a well with an initial speed of 5.00 m/s. The
sound of splash is heard 1.50 s later. Find the depth of the well if the air temperature is
15.0 ˚C?
The speed of the sound is
v = 331√((273+15)/273) = 340. m/s
Let t1 be the flight time of the stone and t2 be the time for the sound wave to reach the
top of the well, then t1+t2 = 1.5 s. We express the depth of the well in two ways and solve
the equation for t1.
d = v0t1+(1/2)gt12 = vt2 ⇒ (1/2)gt12+v0t1 = v(1.5-t1)
⇒ 4.9t1²+345t1-510 = 0 ⇒ t1 = {-345+√(345²+4×4.9×510)}/9.8 = 1.45 s
Plugging it in the first expression of d, we have
d = v0t1+(1/2)gt12 = 5×1.45+(1/2)×9.8×1.45² = 17.6 m
*3. (5pts) A bat flying at 6.00 m/s emits a chirp at 40.0 kHz. If this sound pulse is
reflected by an insect moving away from the bat at 2.00 m/s, what is the frequency of the
echo received by the bat? Assume the air temperature is 25.0 ˚C.
The speed of sound is v = 331√((273+25)/273) = 346 m/s.
We calculate the frequency of the echo reflected by the insect, since the observer (the
insect) is moving away from the source (the bat) and the source is approaching the
observer, we have
f ’ = f(v+vo)/(v-vs) = f(v+(-vi))/(v-vb) = 40×(346-2)/(346-6) = 40.47 kHz
The insect reflects the sound of this frequency, and now the source is the insect and the
observer is the bat. Since the observer is approaching and the source is receding, we
have
fo = f ’ (v+vo)/(v-vs) = f ’(v+vb)/(v-(-vi)) = 40.47(346+6)/(346+2) = 40.94 kHz
***4. (5pts) A uniform string fixed at one end passes over a pulley and then is attached
to a hanging object. The string is horizontal between the fixed end and the pulley. When
you let the hanging object oscillate in the vertical plane, the fundamental frequency of
the horizontal part of the string also oscillates. If the sound of the string changes by one
octave during the oscillation, what maximum angle does the oscillating part of the
string make to the vertical? (Hint: When a frequency is doubled, the sound changes by
an octave.)
Since the sound changes by an octave, the maximum fundamental frequency is twice
the minimum frequency. Thus,
2 = fmax/fmin = (1/2L)√(Fmax/μ)/[(1/2L)√(Fmin/μ)] = √(Fmax/Fmin) ⇒ Fmax = 4Fmin
Thus all we have to do is to calculate the tension. Using Newton’s 2nd law and the
energy conservation equation to the hanging object, we have
mgr(1-cosθmax) = (1/2)mv²+mgr(1-cosθ) ⇒ v² = 2gr(cosθ-cosθmax)
m(v²/r) = F-mgcosθ ⇒ F = mgcosθ+(m/r)×2gr(cosθ-cosθmax) = 3mgcosθ-2mgcosθmax
Thus the tension is maximum at the bottom and minimum at the top. We get
Fmax = 3mg-2mgcosθmax & Fmin = 3mgcosθmax-2mgcosθmax = mgcosθmax
Fmax = 4Fmin ⇒ 3mg-2mgcosθmax = 4 mgcosθmax ⇒ cosθmax = 1/2 ⇒ θmax = 60˚