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Exercises on Oscillations and Waves
Exercise 1. You find a spring in the laboratory. When you hang 100 grams at the end of the spring it stretches 10 cm. You pull the 100 gram mass 6 cm from its equilibrium position and let it go at t = 0. Find an equation for the position of the mass as a function of time t.
Lets first find the period of the oscillations, then we can obtain an equation for
the motion. The period T = 2π
√ m/k. The mass m is 0.1 Kg. To find k, we use the fact that 100 grams causes the spring to stretch an additional 10 cm. Since F = k∆x, we have
mg = k∆x 0 .1(9.8) = k(0.1) k = 9. 8 N/m
The period of the motion is therefore T = 2π
√
- 1 / 9. 8 ≈ 0. 635 sec. At t = 0 the
mass is at its maximum distance from the origin. Thus, x(t) = 0. 6 cos(2πt/T ). Using T = 0.635 sec gives
x(t) = 0. 6 cos(2πt/ 0 .635) x(t) ≈ 0. 6 cos(9. 9 t)
The cosine function is the appropriate one, since at t = 0 the mass is at its maximum distance from equilibrium.
Exercise 1. Joan has two pendula, one has a length of 1 meter, and the other one is longer. She sets them both swinging at the same time. After the 1 meter pendulum has com- pleted 12 oscillations, the longer one has only completed 11. How long is the longer pendulum?
One can solve this problem by taking the ratio of the equation for the periods of the two pendula. Since the period of a pendulum (for small amplitudes) is approximately
T = 2π
√ l/g, we have
T 1 = 2 π
√ l 1 g
T 2 = 2 π
√ l 2 g
for each of the pendula. Dividing these two equations we have
T 1 T 2
√ l 1 l 2
Solving for l 2 we have
l 2 = l 1 (
T 2
T 1
)^2 (2)
Using l 1 = 1 meter, and T 2 /T 1 = 12/11 yields
l 2 = 1(12/11)^2 ≈ 1. 19 meters (3)
So the longer pendulum is 1.19 meters long.
Exercise 1. A spring is hanging freely from the ceiling. You attach an object to the end of the spring and let the object go. It falls down a distance 49 cm and comes back up to where it started. It continues to oscillate in simple harmonic motion going up and down a total distance of 49 cm from top to bottom. What is the period of the simple harmonic motion?
You might think that there is not enough information to solve the problem, but lets try and see if we need to know more about the system. Let the mass of the object be m and the spring constant be labeled k. The period of simple harmonic motion for an ideal spring is given by
T = 2π
√ m k
To solve for the period T , we need to know the ratio of m/k. What does the 49 cm tell us? This is the total distance from the top to the bottom of the simple harmonic motion. Let d = 49 cm. This means that the equilibrium position lies d/2 from the
The force the spring exerts on the rod equals k times the extension of the spring. If θ = 0 is the equilibrium position, then the spring exerts a force equal to k(L/2)θ, where θ is the angle the rod is rotated from its equilibrium position. Putting the pieces together, the equation of motion for the rod is
Iα = τnet
(
M L^2
d^2 θ dt^2
= −rF
M L^2
d^2 θ dt^2
L
k(L
θ 2
The minus sign is needed since the force is a restoring force. If θ is positive, the force (and the torque) is in the negative direction. If θ is negative, the force is in the positive direction. The spring always pushes or pulls the rod back to its equilibrium position. Simplifying the equation above gives
d^2 θ dt^2
3 k M
θ (7)
since the√ L’s cancel. The solution to this equation is θ(t) = A sin(ωt), where ω =
3 k/M. So the period of the motion is T = 2π/ω or
T = 2π
√ M 3 k
Exercise 1. Consider a spring that is standing on end in the vertical position. You place 100 grams on the spring and it compresses a distance of 9.8 cm.
a) If an additional 200 grams are placed on top of the 100 gram mass, how much will the spring compress?
This is an easy question if we assume that the spring obeys Hooke’s Law. If 100 grams compresses the spring 9.8 cm, then 200 grams will compress the spring 2(9.8) = 19.6 cm.
b) What is the spring constant?
The spring constant is defined in the equation Fx = −kx. Consider the 100 gram mass. It exerts a force of F = mg = (0. 1 kg)(9. 8 m/s^2 ) = 0. 98 N on the spring. The compression is 0. 098 m, so the spring constant is
k =
F
x
= 10 N/m (9)
You would get the same result if you considered the 200 gram mass and its compres- sion.
Exercise 1. A massless spring attached to a wall lies on a frictionless table. It has a block of mass 2 kg attached to one end. Initially the block is at rest. Another block, also of mass 2 kg is sliding along on the tabletop with a speed of 8 m/s. At time t = 0, the moving block collides with the block on the spring. The two stick together and oscillate back and forth. If the spring constant is 16 N/m, find an expression x(t) which describes the motion of the two blocks that are stuck together.
Since there are no external forces during the collision, linear momentum is con- served. Let vf be the final speed after the collision. then
f inal total momentum = initial total momentum (m 1 + m 2 )vf = m 1 v 0 (2 + 2)vf = 2(8) vf = 4 m/s
This speed of 4 m/s is the initial speed for the oscillatory motion. Since the spring obeys Hooke’s law, the motion is one of simple harmonic (i.e. sinusoidal) with√ ω =
k/m =
√ 16 /4 = 2 s−^1. The general expression for simple harmonic motion is:
x(t) = x 0 cos(ωt) +
v 0 ω
sin(ωt) (10)
For our example, x 0 = 0 since the blocks are at x = 0 at t = 0. Thus only the sin term is present:
x(t) = 2sin(2t) meters (11)
where t is in seconds.
Exercise 1. An object of mass m is oscillating back and forth in simple harmonic motion. The maximum distance from equilibrium is A, and the period of oscillation is T. At t = 0 the object is at the origin, x = 0, and moving in the −x direction. Find the following in terms of m, T , and A:
a) The equation of motion of the object.
The general form for the equation of motion is x(t) = x 0 cos(ωt) + (v 0 /ω)sinωt. In our case, x 0 = 0. Since the maximum amplitude is A, we have |v 0 /ω| = A. Since the object is initially moving in the ”-” direction, v 0 < 0 so v 0 /ω = −A. Since ω = (2π)/T , the equation of motion is therefore:
x(t) = −Asin(
2 π T
t) (12)
b) The maximum speed of the object.
To find the velocity of the object for any time t, we can just differentiate x(t). Thus,
v(t) =
dx dt v(t) = −
A 2 π T
cos(
2 π T
t)
using the chain rule. Since the cos function varies between −1 and +1, the maximum speed is (2πA)/T.
c) The maximum acceleration of the object.
To find the acceleration of the object for any time t, we can just differentiate v(t):
a(t) =
dv dt a(t) = (
A 2 π T
)^2 sin(
2 π T
t)
using the chain rule again. Since the sin function varies between −1 and +1, the maximum acceleration is ((2πA)/T )^2.
d) The total energy of the object.
The total energy of an object moving in simple harmonic motion equals its kinetic energy as it passes through the equilibrium position. This is true since at x = 0 the object has no potential energy (i.e. it is at a minimum) and all of its energy is in the form of kinetic energy. Thus, we have
Etot =
m 2
v^2 max
=
m 2
2 πA T
)^2
Etot =
2 π^2 mA^2 T 2
Note: all our answers for this problem apply to any type of simple harmonic motion. The answers don’t only apply to a mass on a spring.
Exercise 1. Consider the U-shaped tube containing a fluid shown in the figure. The cross sec- tional area of the tube is A, and the total length of the tube is l. The fluid is pushed up on one side and released. The fluid sloshes back and forth in periodic motion (assuming there are no frictional forces). Is the motion simple harmonic? If so, what is the period of the motion?
To determine what kind of motion will occur, we displace the fluid a distance y from equilibrium and determine what the net force is. Suppose the surface on the right side of the tube is raised a distance y. Then there is a total height difference of 2y above the surface of the left side of the tube. The net force on the fluid will be the weight of the fluid contained in a height of 2y, which we call ∆m. Since the cross-sectional area of the tube is A, the weight unbalance or net force equals Fnet = (∆m)g = (2y)Aρg, where ρ is the density of the fluid. Since the force acts in the opposite direction to y we have:
Fnet = − 2 Aρgy (13)
The net force equals the total mass of the fluid times its acceleration:
m
d^2 y dt^2
= − 2 Aρgy (14)
where m is the total mass of the fluid. Since the total length of the fluid is l, m = Alρ. Subsituting this expression for m into the equation above yields:
Alρ
d^2 y dt^2
= − 2 Aρgy
d^2 y dt^2
2 g l
y
Thus, we see that the restoring force (i.e. or acceleration) is proportional to the displacement from equilibrium (y). The solution to the differential equation above is
a perfect sinusoidal y(t) = Asin(ωt) where ω =
√ 2 g/l. The force is a linear restoring
force and will produce simple harmonic motion. The period T equals 2π/ω:
T = 2π
√ l 2 g
Exercise 2. Eric has a semi-infinite string and is holding it at one end. He moves the end of the string up and down in a sinusoidal motion in time at a rate of 50 times per second. The distance from equilibrium that he moves the end is 0.05 meters. The speed of the wave on the string is 200; m/s. Find an equation for y(x, t) that describes the displacement of the string from equilibrium in terms of x (the distance from the end) and t time.
Since he is moving the end in a perfect sinusoidal motion, the shape of the wave will be a sine wave. The wavelength λ of the traveling sine wave equals λ = v/f = 200 /50 = 4 meters. So an equation for the wave can be
y(x, t) = Asin(
2 π λ
x − 2 πf t)
= 0. 05 sin(
π 2
x − 100 πt) meters
This is a sine wave that travels in the +x direction. If the wave were to travel in the −x direction, the function would be y(x, t) = 0. 05 sin(π 2 x + 100πt) meters. We could also add a phase in the sin argument. Note: if he were to move the end up and down in a motion other than sinusoidal,
then the shape would not be a perfect sine wave.
Exercise 2. Eric takes the same string with the same tension as in Exercise 2.1, but cuts it so its length is 1 meters. He fixes both ends, and forms a standing wave with an amplitude of 0.05 meters. The standing wave produced is the fundamental. Find an equation for y(x, t) that describes the displacement of the string from equilibrium in terms of x and t.
Since the tension is the same as before, the speed of the wave on the string is 200 m/s. The standing wave is the fundamental, which means that there is one anti-mode. Therefore, the wavelength is equal to twice the length of the string, λ = 2L = 2 meters. The frequency of the oscillations is f = v/λ = 200/2 = 100 Hz. For a standing wave, the space and time functions separate. Since the frequency is 100 Hz, the time function is sin(2π 100 t) = sin(200πt). Since the wavelength is 2 meters, the space function is sin(2πx/λ) = sin(πx). Therefore, the function y(x, t) becomes:
y(x, t) = 0. 05 sin(πx) sin(200πt) (16) for
0 ≤ x ≤ 1 meter (17)
The equation is only valid for values of x between 0 and 1 meter, which is the length of the string. We could add a phase in the time part: sin(200πt + φ) which would depend on when we choose t = 0. However, we cannot add a phase to the position part, since x = 0 and x = L = 1 meter are nodes. y(0, t) = y(L, t) = 0 for all times. Also, note the difference between the traveling wave solution of Exercise 2.1 and the standing wave of this exercise.
Exercise 2. Gisela is standing on the side of the autobahn (freeway). Hans has just passed her traveling 44 m/s (100 mph) and is traveling north. Wolfram is traveling towards her from the south at a speed of 60 m/s (136 mph). He is also traveling towards the north. Wolfram honks his horn, which has a frequency of 400 Hz. Take the speed of sound in air to be 340 m/s. Find:
towards a big wall and honks the horn. The horn has a frequency of 400 Hz. She hears the sound that is reflected off the wall, and notices that the two notes, the horn and reflected wave, result in a beat frequency of 4 beats/sec. How fast is she traveling?
At the wall, the note will be heard with a higher frequency than 400 Hz. Since she is traveling towards the wall, she will hear a still higher note. Since the beat frequency is 4 Hz, the note that she hears after the sound has bounced off the wall is (400 + 4) = 404 Hz. It is probably easiest to break the problem up into two parts. First find the frequency that a person standing next to the wall would hear the sound as. Then, take this frequency as the source that Flipper is traveling towards. Since the wall is stationary, vo = 0. Let flippers speed be v. This gives
fat wall = 400(
1 − v/ 340
In this case, Flipper’s horn is the source of the sound. The wall is the observer and is stationary. There is a minus sign in the denominator since she is moving towards the wall (i.e. the frequency at the wall is greater than 400 Hz.) The frequency fat wall is the frequency in air reflected back from the wall. Flipper is driving towards the wall, so she will hear the higher frequency of 404. Thus, we have
404 = fat wall(1 + v/340)
404 = 400(
1 − v/ 340
)(1 + v/340)
1 + v/ 340 1 − v/ 340
We can solve the last equation for v, which gives v ≈ 1 .69 m/s. This is the method that dolphins use to determine if they are closing in on their prey.
Exercise 2. John has two identical guitar strings of the same length. One has a tension F 1. When he plucks this string it produces a sound of 440 Hz for its fundamental note. The second string has a tension that is 1 % higher than the first, F 2 = 1. 01 F 1. If both string are plucked at the same time, what is the beat frequency?
Ir^2 =
P
4 π
Since the power is the same no matter where Hector sits, we must have
I 2 r^22 = I 1 r^21 (21)
where position 1 is at 2 meters, and position 2 is far away. Substuting in the appro- priate quantities, we have
10 −^3 r 22 = (1)2^2 r 2 = 2
r 2 ≈ 63 meters
So Hector should move 63 meters away from the speaker. The actual power P that the speaker is putting out is P = I(4πr^2 ) = 1(4π 22 ) ≈ 50 Watts.
noindent Exercise 2. Lorraine has two organ pipes. Both pipes are open at both ends. One pipe is 60 cm long, and the other is 61 cm long. If both are played in their fundamental mode at the same time, what is the beat frequency? Take the speed of sound in air to be 340 m/s.
The beat frequency is the difference in frequency of the two pipes. Let’s find the fundamental frequency of each pipe. Since the pipes are open at both ends, the fundamental frequency will have the length equal to 1/2 the wavelength. Thus, the fundamental frequency f is related to the length L by
f =
v 2 L
where v is the speed of sound. The first pipe has a frequency of f 1 = 340/(2(0.6)) ≈ 283 .3 Hz. The frequency of the second pipe is f 2 = 340/(2(0.61)) ≈ 278 .7 Hz. So the beat frequency is fbeat = f 1 − f 2 ≈ 283. 3 − 278. 7 ≈ 4 .6 Hz. This means that the sound will have around 4.6 loud-soft oscillations per second.
Exercise 2. Eva notices that the ”A” string on her violin is a little flat. She measures the fre- quency to be 436 Hz. The ”A” note should be 440 Hz. The tension in the string is F 1. To obtain the proper note, she needs to increase the tension to F 2. What is the proper tension F 2 in terms of F 1?
We need to determine how the fundamental frequency depends on tension F. If L is the length of the string, then f = v/(2L). The speed v depends on the tension
according to v =
√ F/μ. Substituting into the equation for f , we have
f =
2 L
√ F μ
Since the length of the string L and the mass per length μ are not changed, we see that the fundamental frequency is proportional to the square-root of F. Taking the ratio of f 2 /f 1 , L and μ cancel giving:
f 2 f 1
√ F 2 F 1
Solving for F 2 , we have
F 2 = F 1 (
f 2 f 1
)^2
= F 1 (
)^2
F 2 ≈ 1. 02 F 1
So Eva needs to increase the tension in the string by around 2%.
Exercise 2. My friend Soumya has given me an exotic wind instrument from the villages of Ben- gal. I am not sure if it will behave as an open-open pipe, or an open-closed pipe. I make overtone notes, and produce successive resonant frequencies of 510, 850, and 1190 Hz. Is the instrument an open-open pipe or an open-closed pipe? What is the fundamental frequency?
If the instrument is an open-open pipe, the resonant frequencies are
So the frequency of the sine wave tone is f = 340/ 0. 708 ≈ 480 Hz.
Exercise 2. Sally’s roommate likes to listen to pure sine wave tones. She sets up two speakers at each end of the room. The speakers are 10 meters apart and are each producing the exact same wave coherently that is a single sine wave of 170 Hz. Sally likes to study in the middle of the room, which is 5 meters from each speaker. However, the constructive interference at this point results in a very loud sound and Sally cannot study well. How far towards one of the speakers should she move so that the waves from the speakers cancel, allowing her to study better? Take the speed of sound in air to be 340 m/s.
To have the sound waves produce destructive interference, the distance from Sally to one speaker minus her distance to the other speaker must be one-half a wavelength. The wavelength of the note is λ = 340/170 = 2 meters. Suppose Sally moves a distance x to the right as shown in the figure. Then the distance to speaker ”1” is d 1 = 5 + x, and the distance to speaker ”2” is d 2 = 5 − x. For destructive interference, d 1 − d 2 must equal λ/2:
λ 2
= d 1 − d 2 2 2
= (5 + x) − (5 − x) 1 = 2 x x =
meter
So if she moves 0.5 meters to the right, she will be 5.5 meters from the speaker on the left, and 4.5 meters from the speaker on the right. The difference in the distance from the speakers is (5. 5 − 4 .5) or 1 meter. This difference is one half a wavelength, which results in the cancelation of the two waves. Now she can study her physics better.
Exercise 2. Consider the set-up shown in the figure. There are two sources of sound waves that are producing the exact same sound wave (i.e. coherent)that is a single sine func- tion of frequency 340 Hz. The speakers are separated by a distance of 20 cm. The
