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The concepts of transverse speed, acceleration, and energy transfer in the context of sine waves traveling along a stretched string. It covers the relationship between wave speed, frequency, and wavelength, as well as the maximum transverse speed and acceleration of a point on the string. The document also discusses the rate of energy transfer in a sine wave and how it can be calculated.
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Typology: Schemes and Mind Maps
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Lana Sheridan
De Anza College
May 21, 2020
Quick Quiz 16.2^1 A sinusoidal wave of frequency f is traveling along a stretched string. The string is brought to rest, and a second traveling wave of frequency 2f is established on the string.
What is the wavelength of the second wave?
(A) twice that of the first wave (B) half that of the first wave (C) the same as that of the first wave (D) impossible to determine
(^1) Serway & Jewett, page 489.
Quick Quiz 16.2^1 A sinusoidal wave of frequency f is traveling along a stretched string. The string is brought to rest, and a second traveling wave of frequency 2f is established on the string.
What is the amplitude of the second wave?
(A) twice that of the first wave (B) half that of the first wave (C) the same as that of the first wave (D) impossible to determine
(^1) Serway & Jewett, page 489.
Consider a point, P, on a string carrying a sine wave.
Suppose that point is at a fixed horizontal position x = 5 λ/4, a constant.
The y coordinate of P varies as:
y
5 λ 4
, t
= A sin(−ωt + 5 π/ 2 )
= A cos(ωt)
The point is in simple harmonic motion!
Sinuso In Figu and dow with an sents sn of the b as that a element frequen element with a sp If we shown i
We can ment at coordin with the
P
t = 0
t = T
A
P
P
P
l
4
1
t = (^) 21 T
t = (^) 43 T
a
b
c
d
x
y
Figure 16.10 One method for
The transverse speed vy is the speed at which a single point on the medium (string) travels perpendicular to the propagation direction of the wave.
We can find this from the wave function
y (x, t) = A sin(kx − ωt)
vy = −ωA cos(kx − ωt) ay = −ω^2 A sin(kx − ωt) = −ω^2 y If we fix x =const. these are exactly the equations we had for SHM!
The maximum transverse speed of a point P on the string is when it passes through its equilibrium position.
vy ,max = ωA
The maximum magnitude of acceleration occurs when y = A (or max value, including sign when y = −A).
ay = ω^2 A
Can a wave on a string move with a wave speed that is greater than the maximum transverse speed vy ,max of an element of the string?
(A) yes (B) no
Can the wave speed be equal to the maximum element speed?
(A) yes (B) no
Can the wave speed be less than vy ,max?
(A) yes (B) no
Waves do transmit energy.
A wave pulse causes the mass at each point of the string to displace from its equilibrium point.
At what rate does this transfer happen? (Find dEdt )
Waves do transmit energy.
A wave pulse causes the mass at each point of the string to displace from its equilibrium point.
At what rate does this transfer happen? (Find dEdt )
Consider the kinetic and potential energies in a small length of string.
Kinetic: dK =
(dm)v (^) y^2 Replacing vy :
dK =
(dm)A^2 ω^2 cos^2 (kx − ωt)
ds − dx =
∂y ∂x
dx
dU =
∂y ∂x
dx
T (Ak cos(kx − ωt))^2 dx
=
μω^2 A^2 cos^2 (kx − ωt) dx
having used v = ω/k and v =
T /μ in the last line.
dK =
μ dx A^2 ω^2 cos^2 (kx − ωt)
dU =
μA^2 ω^2 cos^2 (kx − ωt) dx
Adding dU + dK gives
dE = μω^2 A^2 cos^2 (kx − ωt) dx
Integrating over one wavelength gives the energy per wavelength:
Eλ = μω^2 A^2
∫ (^) λ
0
cos^2 (kx − ωt) dx
= μω^2 A^2
λ 2