Problem solutions

7 A point on the string a distance $y$ from the bottom must have tension sufficient to support all the weight of the string below that point, which is

W (y) = μ y g, T (t) = W (y) = μ y g

v (y) = \sqrt{\frac{T (y)}{μ}} = \sqrt{y g}

A pulse requires time $d t$ to travel a distance $d y$ on the string;

v (y) = \frac{d y}{d t}, d t = \frac{d y}{v (y)} = \frac{d y}{\sqrt{y g}}, T = \int_{0}^{L} \frac{d y}{\sqrt{y g}} = 2 \sqrt{\frac{L}{g}}

8 Use the trig identity

cos A + cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2})

with $A = (π (\frac{x}{1 m} + \frac{8}{s} t)$ , $B = (π (\frac{x}{1 m} - \frac{8}{s} t)$ to get

y_{1} (x, t) + y_{2} (x, t) = 2 cos (π \frac{x}{1.0 m}) cos (π \frac{8 t}{s})

There are always nodes where

π \frac{x_{n}}{1.0 m} = (n + \frac{1}{2}) π, x_{n} = (n + \frac{1}{2}) m = 0.5 m, 1.5 m, . . .

9To the left of the junction

v = \sqrt{\frac{100 N}{0.01 \frac{k g}{m}}} = 100 \frac{m}{s}

On the right side

v^{'} = \sqrt{\frac{100 N}{0.04 \frac{k g}{m}}} = 50 \frac{m}{s}

then

A_{r e f} = \frac{50 - 100}{50 + 100} 0.006 m = - 0.002, A_{t r a n s} = \frac{2 (50)}{50 + 100} 0.006 m = 0.004 m

(the reﬂected wave is inverted). The reﬂected wave has width $u_{r e f} = 0.004 m$ like the incident, the transmitted wave-width is $u_{t r a n s} = \frac{50}{100} 0.004 m = 0.002 m$ . You can see this in the ﬁgure in the notes.

10 The reﬂected wavelength equals the incident ( $1.0 m$ ), the transmitted wavelength is $λ_{t r a n s} = \frac{50}{100} λ_{i n c} = 0.5 m$ .

11 Start with

y (x, t) = A e^{- \frac{{(x - v t)}^{2}}{u^{2}}}

then the velocity of the medium is

v_{y} (x, t) = \frac{d y}{d t} = \frac{2 A v}{u^{2}} (x - v t) e^{- \frac{{(x - v t)}^{2}}{u^{2}}}

and the transverse string-tension is

T_{y} \approx T \frac{d y}{d x} = - \frac{2 A T}{u^{2}} (x - v t) e^{- \frac{{(x - v t)}^{2}}{u^{2}}}

and so the power function for the wave is

℘ (x, t) = - v_{y} (x, t) T_{y} = \frac{2^{2} A^{2} T v}{u^{4}} {(x - v t)}^{2} e^{- \frac{2 {(x - v t)}^{2}}{u^{2}}}

and the total energy carried is

E = \int_{- \infty}^{\infty} ℘ (x, t) d t = \int_{- \infty}^{\infty} \frac{2^{2} A^{2} T v}{u^{4}} {(x - v t)}^{2} e^{- \frac{2 {(x - v t)}^{2}}{u^{2}}} d t

E = \int_{- \infty}^{\infty} \frac{2^{2} A^{2} T}{u^{4}} {(x - v t)}^{2} e^{- \frac{2 {(x - v t)}^{2}}{u^{2}}} v d t = \int_{- \infty}^{\infty} \frac{2^{2} A^{2} T}{u^{4}} z^{2} e^{- 2 \frac{z^{2}}{u^{2}}} d u, u = x - v t

I (a) = \int_{- \infty}^{\infty} e^{- a x^{2}} d x = \sqrt{\frac{π}{a}}

then

\int_{- \infty}^{\infty} x^{2} e^{- a x^{2}} d x = - \frac{d}{d a} I (a) = \frac{1}{2} \sqrt{\frac{π}{a^{3}}}

(differentiate with respect to $a$ ), and so

E = \frac{2^{2} A^{2} T}{u^{4}} \frac{1}{2} \sqrt{\frac{π}{2}} \frac{u^{3}}{2} = \frac{A^{2} T}{u} \sqrt{\frac{π}{2}}

12This is very simple for wavetrains and power

℘_{t r a n s} + ℘_{r e f} = \frac{4 Z Z^{'}}{{(Z + Z^{'})}^{2}} ℘_{I n c} + \frac{{(Z - Z^{'})}^{2}}{{(Z + Z^{'})}^{2}} ℘_{I n c} = \frac{Z^{2} + {Z^{'}}^{2} - 2 Z Z^{'} + 4 Z Z^{'}}{{(Z + Z^{'})}^{2}} ℘_{i n c} = \frac{{(Z + Z^{'})}^{2}}{{(Z + Z^{'})}^{2}} ℘_{i n c} = ℘_{i n c}

For wave pulses, you must work a little harder.

E_{i n c} = \frac{A^{2} T}{u} \sqrt{\frac{π}{2}}, E_{r e f} = (\frac{v^{'} - v}{v^{'} + v})^{2} \frac{A^{2} T}{u} \sqrt{\frac{π}{2}}, E_{t r a n s} = (\frac{2 v^{'}}{v^{'} + v})^{2} \frac{A^{2} T}{u \frac{v^{'}}{v}} \sqrt{\frac{π}{2}}

so check

E_{r e f} + E_{t r a n s} = (\frac{v^{'} - v}{v^{'} + v})^{2} \frac{A^{2} T}{u} \sqrt{\frac{π}{2}} + (\frac{2 v^{'}}{v^{'} + v})^{2} \frac{A^{2} T}{u \frac{v^{'}}{v}} \sqrt{\frac{π}{2}} = \frac{A^{2} T}{u} \sqrt{\frac{π}{2}} ((\frac{v^{'} - v}{v^{'} + v})^{2} + \frac{4 {v^{'}}^{2} v}{v^{'} {(v^{'} + v)}^{2}})

= \frac{A^{2} T}{u} \sqrt{\frac{π}{2}} (\frac{{v^{'}}^{2} + v^{2} - 2 v^{'} v + 4 v^{'} v}{{(v^{'} + v)}^{2}}) = \frac{A^{2} T}{u} \sqrt{\frac{π}{2}} (\frac{{v^{'}}^{2} + v^{2} + 2 v v^{'}}{{(v^{'} + v)}^{2}}) = \frac{A^{2} T}{u} \sqrt{\frac{π}{2}} = E_{i n c}

13 There must be a node at the end where it is attached to the car, and an antinode at or near the free end. The longest wave is therefore $1 \cdot \frac{λ}{4} = 1 m$ , $λ = 4 m$ , the next is $\frac{λ}{2} + \frac{λ}{4} = 1 m$ , $λ = \frac{4}{3} m$ , and so forth.

14 Take a wave-train such as $y (x, t) = A sin (k x - ω t)$ on a string and compute the kinetic energy of a section of width $d x$ located at $x$ ; this is easy to do from the displacement function and ﬁrst principles

d K E = \frac{1}{2} d m {(\frac{d y}{d t})}^{2} = \frac{1}{2} (μ d x) (- A ω cos (k x - ω t))^{2}

Now determine the total kinetic energy stored between $x = 0$ and $x = λ = \frac{2 π}{k}$ ;

K E = \int_{0}^{λ} \frac{1}{2} (μ d x) (- A ω cos (k x - ω t))^{2} = \frac{1}{2} A^{2} ω^{2} μ \frac{λ}{2}

Divide this by the average power carried by the wave, this eliminates a lot of things and so the two quantities can be readily compared;

\frac{K E}{\bar{℘}} = \frac{\frac{1}{2} λ}{v} = \frac{1}{2} f = \frac{1}{2 T}

rearrange

\frac{K E}{T} = \frac{1}{2} \bar{℘}

The left hand side is the energy within one wavelength divided by the time it takes one wavelength of the wave to pass a given point. This is the rate of kinetic energy ﬂow, but $℘$ is the rate of total energy ﬂow, kinetic and potential, and so we learn that the kinetic and potential energies trapped within a wavelength of the wave are equal, and

K E = P E = \frac{1}{2} T \bar{℘}

This is characteristic of elastic materials, and therefore if the results are true for a wave-train, they will also be true of a wave-pulse. It is not difficult to show that the wave-pulse is a superposition of wave-trains.