1 Problem solutions

2Let $y (x, t) = A sin (k x - ω t)$ , with $k = \frac{2 π}{λ}, ω = 2 π f$ , then the medium moves at

v_{y} (x, t) = \frac{d}{d t} y (x, t) = A ω cos (k x - ω t)

and so

- A ω \leq v_{y} (x, t) \leq A ω, A ω = 2 π A f

Acceleration of medium;

a_{y} (x, t) = \frac{d}{d t} v (x, t) = - A ω^{2} sin (k x - ω t)

and so

- A ω^{2} \leq a_{y} (x, t) \leq A ω^{2}, A ω = 2 π A f

3 Compare to

y (x, t) = A sin (k x - ω t)

you see that

A = 0.002 m, k = \frac{2 π}{λ} = \frac{20}{m}, ω = 2 π f = \frac{600}{s}

λ = 0.31415 m, f = 95.49 H z, v = c = f λ = 30 \frac{m}{s}

and

v_{y, m a x} = A ω = (0.002 m) (600 \frac{1}{s}) = 1.2 \frac{m}{s}

4 $T = \frac{1}{f} = 0.2 s$ , $f = 5 H z$ , $v = c = f λ = (5 H z) (0.05 m) = 0.25 \frac{m}{s}$

y (x, t) = 0.05 m sin (\frac{2 π}{0.05 m} (- x - 0.25 \frac{m}{s} t))

v_{1} = 170 \frac{m}{s} = \sqrt{\frac{120 N}{μ}}, v_{2} = 180 \frac{m}{s} = \sqrt{\frac{T}{μ}}

square and divide

\frac{{(180)}^{2}}{{(170)}^{2}} = \frac{T}{120 N}, T = 134.53 N

6 $A = 0.002 m$ , $k = \frac{20}{m}$ , $ω = \frac{600}{s}$ so $v = c = f λ = \frac{ω}{k} = 30 \frac{m}{s}$ , then

30 \frac{m}{s} = \sqrt{\frac{15 N}{μ}}, μ = 0.01666 \frac{k g}{m}

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}}