Sound and Wave - Wave phenomena

A string consists of two parts attached at x = 0. The right part of the string (x  0) has mass μ_r per unit length and the left part of the string (x  0) has mass  mass μ_l per unit length. The string tension is T. If a wave of unit amplitude travels along the left part of the string, as shown in the figure above, what is the amplitude of the wave that is transmitted to the right part of the string?

A. 1
B. 

C. 

D. 

E. 0

(GR9677 #80)

Solution:

A. FALSE
Amplitude = 1, if  μ_l  = μ_r  but from the picture  μ_l ≠ μ_r
Since μ_l  μ_r suggests we expect  the transmitted amplitude will be less than 1

B. FALSE
The answer suggests that the transmitted amplitude will be bigger than 1, it should be less than 1
Let μ_l = 1, μ_r = 4
A = 2 / (1 + √¼) = 2/(3/2) = 4/3

C. TRUE
The answer suggests that the transmitted amplitude will be less than 1
Let μ_l = 1, μ_r = 4
A = 2(√¼) / (1 + √¼) = 1/(3/2) = 2/3

D. FALSE
The answer suggests that A = 0, if  μ_l  = μ_r

E. FALSE
Amplitude = 0 if  μ_r → ∞

Answer: C 

Sound and Wave - Beats and Tunning

A piano tuner who wishes to tune the note D₂ corresponding to a frequency of 73.416 hertz has tuned A₄  to a frequency of 440.000 hertz. Which harmonic of D₂ (counting the fundamental as the first harmonic) will give the lowest number of beats per second, and approximately how many beats will this be when the two notes are tuned properly?

	Harmonic		Number of Beats
A.	6		5
B.	6		0.5
C.	5		0.1
D.	3		0.372
E.	2		4.5

(GR9677 #81)

Solution:

Beats are the alternating constructive and destructive interference produced when two sound waves of different frequency approaching our ear. If there is no difference in frequency, there will be no beats. To minimize or to get the lowest number of beats, we set the beat frequency to zero.

440.000 − 73.416n = 0
n ≈ 440/73 ≈ 6

Number of beats = |440.000 − (73.416)(6)|  = |440.000 − 440.496| =  0.496 ≈ 0.5

Answer: B 

Sound and Wave - Wave Phenomena

Consider a particle moving without friction on a rippled surface, as shown above. Gravity acts down in the negative h direction. The elevation h(x) of the surface is given by h(x) = d cos (kx). If the particle starts at x = 0 with a speed v in the x direction, for what values of v will the particle stay on the surface at all times?

A.
B.
C.
D.
E. 

(GR9677 #83)

Solution:

Gravity pulls the particle down. 
Particle will stay on the surface if  a_max = g

For a wave, a_max = ω²A = ω²d
(A = d = amplitude)

Velocity of the particle: v =  λ / T 

From the picture: λ = 2π/k

v = 2π/kT  = ω/k
v² = ω²/k²
ω² = v²k²

a_max = v²k²d = g 
v² = g/k²d
v = √(g/k²d)

Answer: D 

Sound and Wave - Standing Wave

Small-amplitude standing waves of wavelength λ occur on a string with tension T, mass per unit μ, and length L. One end of string is fixed and the other end is attached to a ring of mass M that slides on a frictionless rod, as shown in the figure above. When gravity is neglected, which of the following conditions correctly determines the wavelength? (You might want to consider the limiting cases M → 0 and M → ∞.)

A. μ/M = (2π/λ) cot (2πL/λ)
B. μ/M = (2π/λ) tan (2πL/λ)
C. μ/M = (2π/λ) sin (2πL/λ)
D. λ = 2L/n, n = 1, 2, 3, ...
E. λ = 2L/(n + ½), n = 1, 2, 3, ...

(GR9677 #85)

Solution:

Consider the limiting cases M → 0 and M → ∞.

D and E do not depend on M → FALSE

If M → 0 , μ/M → ∞

C. FALSE because sin (2πL/λ)  cannot go to infinity

tan α = sin α / cos α → can go to infinity if cos α = 0

cot α = cos α / sin α → can go to infinity if sin α = 0

If M → ∞ , μ/M → 0, ring will not move = nodes on left and right side
The only possible wavelength: λ = 2L

2πL/λ = 2πL/2L = π

sin π = 0 → cot π = ∞ and tan π = 0 

 

Answer: B 

Classical Mechanics - Wave equation

The equation where A, T, and λ are positive constants, represents a wave whose

A. Amplitude is 2A
B. Velocity is in the negative x–direction
C. Period is T/λ
D. Speed is x/t
E. Speed is λ/T

(GR8677 #04)

Solution:

(A) FALSE.
Amplitude  = maximum displacement = y_max = A.

(B) FALSE.
For a wave traveling to the right (positive x-direction): y = ƒ(x − vt)
(x − vt) = constant.
As t increases, x must increases to keep (x − vt) = constant

For a wave traveling to the left (negative x-direction): y = ƒ(x + vt)
As t increases, x decreases to keep (x + vt) = constant.

The problem gives:  y = ƒ(vt − x)
As t increases, x must increases to keep (x − vt) = constant.
The waves is traveling to the right.

(C) FALSE.
The unit of T/λ (second/meter) does not match with the unit is period (second).

(D) FALSE.
The speed of  the wave is λ/T.

(E) TRUE.
See D.

Answer: E

Sound and Wave - Doppler Effect

A source of 1-kilohertz sound is moving straight toward you at a speed 0.9 times the speed of sound. The frequency you receive is 

A. 0.1 kHz
B. 0.5 kHz
C. 1.1 kHz
D. 1.9 kHz
E. 10 kHz

(GR8677 #12)

Solution:

Moving toward observer
→ f increases
→ f  > 1 kHz
→ A and B are FALSE

Speed 0.9 times the speed of sound
→ f increases greatly
→ E. TRUE

Answer: E

Calculation:





+ sign = receding
− sign = approaching

Sound and Wave - Group Velocity

The dispersion law for a certain type of wave motion is ω = (c²k²+m²)^½, where ω is the angular frequency, k is the magnitude of the propagation vector, c and m are constants. The group velocity of these waves approaches 

A. Infinity as k → 0 and zero as k → ∞
B. Infinity as k → 0 and c as k → ∞
C. c as k → 0 and zero as k → ∞
D. Zero as k → 0 and infinity as k → ∞
E. Zero as k → 0 and c as k → ∞

(GR8677 #59)

Solution:





As k → 0, v_g = 0
As k → ∞, v_g = c
since m ≪ →

Answer: E

Sound and Wave - Frequency

At 20^oC, a pipe open at both ends resonates at a frequency of 440 hertz. At what frequency does the same pipe resonate on a particularly cold day when the speed of sound is 3 percents lower than it would be at 20^oC?

A. 414 Hz
B. 427 Hz
C. 433 Hz
D. 440 Hz
E. 453 Hz

(GR0177 #50)

Solution:



Answer: B