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Unit 1 Waves

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1. An oscillating block–spring system takes 0.75 s to begin repeating its motion.
Find (a) the period, (b) the frequency, and (c) the angular frequency.

(a) Given the time taken to execute one cycle of the motion (period). The period is T = 0.75 s.
(b) Frequency is simply the reciprocal of the period: f = 1/T.
f = 1.3 Hz, where the SI unit Hz stands for Hertz, which means a cycle-per-second.
(c) Since 2π radians are equivalent to a cycle, the angular frequency ω (in radians per second) is related to frequency f by ω = 2πf
ω = 8.4 rad s-1.

2. In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 120 Hz.
Find (a) the amplitude, (b) the maximum blade speed, and (c) the magnitude of the maximum blade acceleration.

(a) The amplitude is half the range of the displacement, or xmax = 1.0 mm.
(b) The maximum speed vmaxmax is related to the amplitude xmax by vmax = ωxmax where ω is the angular frequency. Since ω = 2πf, where f is the frequency,
vmax= ωxmax
vmax= 2π (120Hz) (1.0 X 10-3)
vmax= 0.75 ms-1.
(c) The maximum acceleration is amax= ω2xmax
amax= (2 π f)2xmax
amax= [2 π (120Hz)]2 (1 X10-3m)
amax= 5.7 X102 ms-2.

3. An oscillator consists of a block attached to a spring (k = 400 Nm-1). At some time t, the position (measured from the system’s equilibrium location), velocity, and acceleration of the block are x = 0.100 m, v = 13.6 ms-1, and a = -123 ms-2.
Calculate (a) the frequency of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.

(a)ω2 = (-amax)/( xmax)
ω = √[(-123ms-2)/( 0.100m)]

f = 5.58 Hz. = 5.58 Hz.
(b) ω = √ [k/m]m= k/ω2
m = [400Nm-1/rads s-1]

m = 0.325 kg.

(c) By energy conservation ET = Ep + EkET = ½ kx2 + ½ mv2
At the turning points where X is a maximum Ek = 0, so ET = Ep and when x = 0 and V = vmax Ep = 0 so ET = Ek.
ET = ½ k(xmax)2 + 0
ET = ½ m (vmax)2 + 0
We don’t know vmax but we know x and v.
½ k(xmax)2 = ½ mv2 + ½ kx2
k(xmax)2 = (mv2) + kx2
(xmax)2 = [(mv2)/k] + x2
(xmax)2 = {[(0.325kg)(13.6ms-1) 2]/400Nm-1} + (0.100m)2
xmax = 0.400m

question 44. A wooden block of mass M = 5.4 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant 6000 Nm-1. A bullet of mass m = 9.5 g and horizontal velocity, v, 630 ms-1 strikes and is embedded in the block. Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the velocity of the system immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

(a) The collision is inelastic so kinetic energy is not conserved, but momentum is conserved in all collisions.
Momentum before collision = momentum after collision
MV + mv = (M+ m) vmax where vmax is the velocity of the block and bullet after the collision. As this velocity happens at the equilibrium position, it must be the maximum velocity with which the system will operate.
vmax = [MV + mv]/[M+ m]
vmax = [(5.4 kg)(0 ms-1) + (0.0095 kg)(630ms-1)] / [(5.4 kg) + (0.0095 kg)]
vmax = 1.1 ms-1

(b) By energy conservation ET = Ep + EkET = ½ kx2 + ½ mv2
At the turning points where X is a maximum Ek = 0, so ET = Ep and when x = 0 and V = vmax Ep = 0 so ET = Ek.
ET = ½k(xmax)2 + 0
ET = ½(M+m) (vmax)2 + 0
We know vmax.
½k(xmax)2 = ½(M+m)(vmax)2
(xmax)2 = [(M+m)(vmax)2] / k
(xmax)2 = [(5.4kg) + (9.5 X 10-3 )] [(1.1 ms-1)2] / 6000Nm-1
(xmax) = 3.3 X 10-2 m

Alternative method
vmax= ωxmax

1.1 ms-1= 2π (1/T) (xmax)
1.1 ms-1= (2π) [1/[(2π)((k/m)] (xmax)
(xmax) = 1.1 ms-1 / {(2π) [1/[(2π)((k/m)]}
( xmax) = 1.1 ms-1 [(m/k)] (2π)] / (2π)
(xmax) = 1.1 ms-1 [(m/k)]
(xmax) = 1.1 ms-1 [ (5.4kg + 0.0095kg) / (6000 Nm-1)]
(xmax) = 3.3 X 10-2 m

5. A vertical spring stretches 9.6 cm when a 1.3 kg block is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional 5.0 cm downward and released from rest. Find the (b) period, (c) frequency, (d) amplitude, and (e) maximum speed of the resulting SHM.
(a) According to Hooke’s Law: F = -kx
k = F/x
k = mg/x
k = (1.3kg) (9.8ms2) / (9.6X10-2m)
k = 1.33 X 102 Nm-1
(b) T = 1/f
T = 2π √[m/k]
T = 2π √[(1.3 kg) / (1.33 X 102 Nm-1)]
T = 0.62s
(c) f = 1/T
f = 1/0.62s
f = 1.6Hz
(d) Given that the displacement from equilibrium position is 5.0 cm downward then amplitude is 5.0 cm.
(e) At the maximum extension ET = EPinitial = –mgx + ½kx2 (sum of gravitational potential energy and state potential, gravitational potential energy is negative because the mass will gain energy as it moves up. Consider the displacement of the block as y = displacement due to weight + displacement due to additional extension (ymax = 0.096m + 0.050m).
EPinitial = Epgrav + Epstate
EPinitial = – mgymax + ½k(ymax)2
EPinitial = – (1.3 kg) (9.8ms2) (0.096m + 0.050m) + ½ (1.33 X 102 Nm-1) (0.096m+ 0.050m)2
EPinitial = – 0.44J
At equilibrium position
EPequilibrium = Epgrav + Epstate
EPequilibrium = – mgy + ½ky2
EPequilibrium = – (1.3 kg) (9.8ms-2) (0.096m) + ½(1.33 X 102 Nm-1) (0.096m)2
EPequilibrium = – 0.61J
At the equilibrium position: EPinitial = Ep + Ek
– 0.44J = – 0.61J + ½m(vmax)2
Vmax = √[(– 0.44J) – (– 0.61J) /(1.3kg)]
Vmax = 0.51ms-1.

6. What is the frequency of a simple pendulum 2.0 m long (a) in a room, (b) in an elevator accelerating upward at a rate of 2.0 ms-2 and (c) in free fall?
(a) T = 2π √[l/g]
f = [1/2p] √[g/l]
f = [1/] √[(9.80ms-2/2.0m]
f = 0.35Hz
(b) In the elevator the forces acting on the pendulum are Tension (T) and the force of gravity (mg)
By Newton's second law, T + mg = ma
a = ae + ap were ae is the acceleration of the elevator and ap the acceleration of the pendulum with respect to the elevator. They are in opposite directions.
T + m(g - ae) = map
f = [1/] √[(g - ae)/l]
f = [1/] √[(9.80ms-2 - (-2.0ms-2)/2.0m]
f = 0.39Hz
(c)
In free fall the ae + ap equal in magnitude but opposite in direction so
f = [1/] √[(g - ae)/l]
f = [1/] √[(9.80ms-2 - (9.80.0ms-2)/2.0m/l]
f = 0 Hz (no oscillation)

7. A sinusoidal wave of frequency 500 Hz has a speed of 350 ms-1. (a) How far apart are two points that differ in phase by p/3 rad? (b) What is the phase difference between two displacements at a certain point at times 1.00 ms apart?
Using v = fλ, we find the length of one cycle of the wave is
λ = 350/500 = 0.700 m = 700 mm.
From f = 1/T, we find the time for one cycle of oscillation is T = 1/500
T = 2.00 × 10-3 s =2.00 ms.
T = 2.00 ms
(a) A cycle is equivalent to 2π radians, so that π/3 rad corresponds to one-sixth of a cycle.
The corresponding length, therefore, is λ/6
λ/6 = 700 mm/6
λ/6 = 117 mm.
(b) The interval 1.00 ms is half of T and thus corresponds to half of one cycle, or half of 2π rad. Thus, the phase difference is (1/2)(2π) = π rad.
Phase difference = π rad.

8. A string fixed at both ends is 8.40 m long and has a mass of 0.120 kg. It is subjected to a tension of 96.0 N and set oscillating. (a) What is the speed of the waves on the string? (b) What is the longest possible wavelength for a standing wave? (c) Give the frequency of that wave.
(a) The wave speed is given by v = √[τ/μ] where τ is the tension in the string and μ is the linear mass density of the string.
Since the mass density is the mass per unit length, μ = M/L, where M is the mass of the string and L is its length.
Thus v = √[τL/M]
v = √{[(96.0N)(8.40m)/(0.120kg)]}
v = 82.0 ms-1
(b) The longest possible wavelength λ for a standing wave is related to the length of the string by L = λ/2
λ = 2L
λ = 2(8.40 m)
λ = 16.8 m.
(c) The frequency is f = v/λ
f = (82.0 ms-1) / (16.8 m)
f = 4.88 Hz.

q99. A nylon guitar string has a linear density of 7.20 gm-1 and is under a tension of 150 N. The fixed supports are distance D, 90.0 cm apart. The string is oscillating in a standing wave pattern. Calculate the (a) speed, (b) wavelength, and (c) frequency of the traveling waves whose superposition gives this standing wave.
(a) The wave speed is given by v = √[τ/μ] where τ is the tension in the string and μ is the linear mass density of the string.
Thus v = √[(150N)/(7.20 X 10-3kgm-1)]
v = 1.44 X 102 ms-1
(b) From the diagram, we find the wavelength of the standing wave to be λ = (2/3)(90.0 cm)
λ = 0.60 m.
(c) The frequency is f = v/λ
f = (1.44 X 102 ms-1) / (0.60 m)
f = 240 Hz.

10. Earthquakes generate sound waves inside Earth. Unlike a gas, Earth can experience both transverse (S) and longitudinal (P) sound waves. Typically, the speed of S waves is about 4.5 kms-1, and that of P waves 8.0 kms-1. A seismograph records P and S waves from an earthquake. The first P waves arrive 3.0 min before the first S waves. If the waves travel in a straight line, how far away did the earthquake occur?
(a) If d is the distance from the location of the earthquake to the seismograph and v is the speed of the S waves, then the time for these waves to reach the seismograph is ts = [d/vs] and similarly, the time for P waves to reach the seismograph is tp = [d/vp].
Time delay ΔT = ts - tp
ΔT = [d/vs]- [d/vp]
ΔT = d(vp-vs) / (vsvp)
d = [(vsvp) ΔT] / (vp-vs)
d = {[(4.5 X 103ms-1)(8.0 X 103ms-1)(3.0 X 60s)] / [(8.0 X 103ms-1 - (4.5 X 103ms-1)]}
d = 1.9 106m

q1111. Two point sources S1 and S2 emit sound of wavelength λ = 2.00 m. The emissions are isotropic (the same intensity of radiation in all directions) and in phase, and the separation between the sources, d is 16.0 m. At any point P on the x axis, the wave fromS1 and the wave from S2 interfere. When P is very far away (x ≈ ∞), what are (a) the phase difference between the arriving waves from S1 and S2 and (b) the type of interference they produce? Now move point P along the x axis toward S1. (c) Does the phase difference between the waves increase or decrease? At what distance x do the waves have a phase difference of (d) 0.50λ, (e) 1.00λ, and (f) 1.50λ?
(a) If point P is infinitely far away, then the small distance d between the two sources is of no consequence (they seem effectively to be the same distance away from P). Thus, there is no perceived phase difference.
(b) Since the sources oscillate in phase, then the situation described in part (a) produces fully constructive interference.
(c) By Pythogoras, the distance S2P is √[d2+x2]
So the path length difference for waves traveling from S1 and S2 ΔL = √[d2+x2] - x.
The phase difference in “cycles” (in absolute value) is Δφ = [ΔL / λ]
Δφ = {√[(d2+x2) - x]} / λ
Δφ = {√[(d2+x2) - x]} / λ
Consider ΔL = λ / 2 then √[d2+x2] becomes [x + λ / 2] (the diffrerence between the hypothenuse and x).
[d2+x2] = [x + λ / 2]2
[d2+x2] = x2 + xλ + ( λ2 / 4)
x2 - x2 - xλ= -d2+ (λ2 / 4)
xλ= d2 - (λ2 / 4)
x= (d2 / λ) - (λ / 4)
So in general if ΔL = n λ where n > 0, then x= (d2 / 2 nλ) - (nλ / 2)
x= (d2 / 2nλ) - (nλ / 2)
x= [(16.0 m) 2 / 2n(2.00m)] - (n(2.00m) / 2)
x= {[(256) / 4n] - (n)}
x= {[(64) / n] - n}
From this equation, it is clear that for d = 16.0 m and λ = 2.00 m that as x increases the phase difference between the waves decreases.
(d) For ΔL = 0.50λ, x = [(64.0 / 0.50) - 0.50]
x = 127.5m
(d) For ΔL = 1.00λ, x = [(64.0 / 1.00) - 1.00]
x = 63.0m
(d) For ΔL = 1.50λ, x = [(64.0 / 1.50) - 1.50]
x = 41.2m

12. A certain sound source is increased in sound level by 30.0 dB. By what multiple is its intensity increased?
Let I1 be the original intensity and I2 be the final intensity.
The original sound level is β1 = (10 dB) log(I1 / I0) and the final sound level is β2 = (10 dB) log(I2 / I0) where I0is the reference intensity (threshold of hearing).
Since β2 = β1 + 30dB
Then (10 dB) log(I2 / I0) = (10 dB) log(I1 / I0) + 30dB
(10 dB) log(I2 / I0) - (10 dB) log(I1 / I0) = 30dB
log(I2 / I0) - log(I1 / I0) = 3
log(I2 / I1) = 3
(I2 / I1) = 103

13.Suppose that the sound level of a conversation is initially at an angry 70 dB and then drops to a soothing 50 dB. Assuming that the frequency of the sound is 500 Hz, determine the (a) initial and (b) final sound intensities.
(a) β1 = (10 dB) log(I1 / I0)
70 dB = (10 dB) log(I1 / 10-12)
7 = log(I1 / 10-12)
7 = log I1 - log 10-12
log I1 = 7 + (-12)
I1 = 1X 10-5 Wm2
(b) β2 = (10 dB) log(I2 / I0)
50 dB = (10 dB) log(I2 / 10-12)
5 = log(I2 / 10-12)
5 = log I2 - log 10-12
log I2 = 5 + (-12)
I2 = 1X 10-7 Wm2

14. (a) Find the speed of waves on a violin string of mass 800 mg and length 22.0 cm if the fundamental frequency is 920 Hz. (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string?
(a) When the string (fixed at both ends) is vibrating at its lowest resonant frequency, exactly one-half of a wavelength fits between the ends.
Thus, λ = 2L.
Now v = fλ
v = 2Lf
v = 2(0.220 m)(920 Hz)
v = = 405 ms-1.
(b) The wave speed is given by v = √[τ/μ] where τ is the tension in the string and μ is the linear mass density of the string.
If M is the mass of the (uniform) string, then μ = M/L.
Thus, τ = μv2
τ = (M/L)v2
τ = [(800 × 106 kg)/(0.220 m)] (405 ms-1)
τ = 596 N.
(c) The wavelength is λ = 2L
λ = 2(0.220 m)
λ = 0.440 m.
(d) The frequency of the sound wave in air is the same as the frequency of oscillation of the string. The wavelength is different because the wave speed is different. If v is the
speed of sound in air, the wavelength in air is
λa = [va / f]
λa = (343 m/s)/(920 Hz)
λa = 0.373 m.

15. The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded together with a tuning fork that is oscillating accurately at concert A (440 Hz).What is the (a) period of the violin string oscillation (b) frequency of the violin string oscillation?
(a) Let the period be T.
Then the beat frequency [1/T - 440Hz] = 4.00 beats s-1.
T = 2.25 × 103s.
(b) f = 1/T
f = 1/[2.25 × 10–3s]
f = 444 Hz
The string that is “too tightly stretched” has the higher tension and thus the higher (fundamental) frequency.
Alternative solution: f - 440 Hz = 4 beats s-1
f = 444 Hz

16. A stationary motion detector sends waves of frequency 0.150 MHz and speed 343 ms-1 toward a truck approaching at a speed of 45.0 ms-1. What is the frequency of the waves (a) detected by the truck, (b) reflected back to the detector?
(a) fo = fs [(v ± vo) / (v ± vs)]
For a stationary source and a moving observer towards the source fo = fs [(v + vo) / (v - vs)]
fo = 1.50 X 105 Hz [(343 ms-1+ 45.0 ms-1) / (343 ms-1 - 0)]
fo = 1.70 X 105 Hz
(b) The reflected wave from the truck now as a moving wave to be reflected back to the stationary detector.
For a moving source towards the observer and a stationary observer fo = fs [(v + vo) / (v - vs)]
fo = 1.70 X 105 [(343 ms-1+ 0) / (343 ms-1 - 45.0 ms-1)]
fo = 1.95 X 105 Hz

17. A siren emitting a sound of frequency 1000 Hz moves away from you toward the face of a cliff at a speed of 10 m-1. Take the speed of sound in air as 330 m-1. (a) What is the frequency of the sound you hear coming directly from the siren? (b) What is the frequency of the sound you hear reflected off the cliff? (c) What is the beat frequency between the two sounds? Is it perceptible (less than 20 Hz)?
(a) For a moving source away from the observer and a stationary observer fo = fs [(v + vo) / (v + vs)]
fo = 1.00 X 103 [(330 ms-1+ 0) / (330 ms-1 + 10.0 ms-1)]
fo = 9.7 X 102 Hz
(b) The reflection from the cliff is the same as a stationary observer and a source moving towards the observer.
fo = fs [(v + vo) / (v - vs)]
fo = 1.00 X 103 [(330 ms-1+ 0) / (330 ms-1 - 10.0 ms-1)]
fo = 1.03 X 103 Hz
Beat frequency = 1.03 X 103 Hz - fo = 9.7 X 102 Hz
Beat frequency = 60 beats s-1 (too large to be perceptible)

     
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