Sine Waves: Transverse Speed, Acceleration, and Energy Transfer, Schemes and Mind Maps of Particle Physics

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Waves

Transverse Speed and Acceleration

Power of a Wave

Lana Sheridan

De Anza College

May 21, 2020

Last time

• solutions to the wave equation
• (^) sine waves (covered in lab)

Question

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.

Question

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.

Sine waves

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

Sine waves: Transverse Speed and Transverse

Acceleration

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)

Sine waves: Transverse Speed and Transverse

Acceleration

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

Questions

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

Questions

Can the wave speed be equal to the maximum element speed?

(A) yes (B) no

Questions

Can the wave speed be less than vy ,max?

(A) yes (B) no

Rate of Energy Transfer in Sine Wave

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 )

Rate of Energy Transfer in Sine Wave

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)

Rate of Energy Transfer in Sine Wave

ds − dx =

∂y ∂x

dx

dU =

T

∂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.

Rate of Energy Transfer in Sine Wave

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