Question

A sinusoidal transverse wave of wavelength $18 cm$ travels along a string in the positive direction of an $x$ -axis. The displacement $y$ of the string particle at $x = 0$ is given in the figure shown below as a function of time $t .$ The scale of the vertical axis is set by $y_{s} = 4.0 cm$ . The wave equation is to be in the form $y (x, t) = y_{m} \sin (k x \pm ω t + ϕ)$ .

(a) At $t = 0$ , is a plot of $y$ versus $x$ in the shape of a positive sine function or a negative sine function?

What are (b) $y_{m}$ , (c) $k$ , (d) $ω$ , (e) $ϕ$ , (f) the sign in front of $ω$ and (g) the speed of the wave?
(h) What is the transverse velocity of the particle at $x = 0$ when $t = 5.0 s$ ?

Accepted Answer

A sinusoidal transverse wave of wavelength $18 cm$ travels along a string in the positive direction of an $x$ -axis. The displacement $y$ of the string particle at $x = 0$ is given in the figure shown below as a function of time $t$ . The scale of the vertical axis is set by $y_{s} = 4.0 cm$ The wave equation is to be in the form $y (x, t) = y_{m} \sin (k x \pm ω t + ϕ)$ .

Since, the wave is traveling in the positive direction, the wave equation will be,

$y (x, t) = y_{m} \sin (k x - ω t + ϕ) . . . (1)$ .

Substituting $x = 0$ in the equation $(1)$ , we get,

$y (0, t) = y_{m} \sin (- ω t + ϕ) . . . (2)$ .

At, $t = 0$ , $y = 0$ . Hence, $ϕ = 0, π, 2 π, . . .$

However, since the displayed sine function of time is positive, $ϕ = π$ .

Since, the wavelength $λ = 18 cm$ , $t = 0$ .

Also, the maximum displacement is $ϕ$ .

We also note that the time period $T = 10 s$ or $ω = \frac{2 π}{T} = \frac{2 π}{10} = 0.2 π$ .

Hence, the wave equation can be written as,

$y (x, t) = (4.0) \sin (\frac{π}{9} x - 0.2 π t + π) . . . (3)$ .

(a) At $t = 0$ , is a plot of $y$ versus $x$ in the shape of a positive sine function or a negative sine function?

Here, the equation reduces to,

$y (x, 0) = (4.0) \sin (\frac{π}{9} x + π) = - (4.0) \sin (\frac{π}{9} x)$ .

So, this is a negative sine function.

What are (b) $y_{m}$ ,

We note from equation $(3)$ that the value is $y_{m} = 4.0 cm$ .

(c) $k$ ,

As shown, $k = \frac{π}{9} = 0.35 {cm}^{- 1}$ .

(d) $ω = 0.2 π = 0.63 s^{- 1}$ ,

(e) $ϕ = π$ ,

(f) the sign in front of $ω,$

We observe from equation $(3)$ that the sign is minus.

and,

(g) the speed of the wave?

Speed of the wave

$v = \frac{λ}{T} = \frac{18}{10} = 1.8 cm s^{- 1}$ .

(h) What is the transverse velocity of the particle at $x = 0$ when $t = 5.0 s$ ?

The transverse velocity of the wave is given by, differentiating equation $(3)$

$\frac{d y}{d t} = (- 0.2)) (4.0) \cos (\frac{π}{9} x - 0.2 π t + π) . . . (4)$ .

Substituting the given values,

$\frac{d y}{d t} = (- 0.2)) (4.0) \cos (\frac{π}{9} \times 0 - 0.2 π \times 5.0 + π) = - 0.8 \cos (- π + π) = - 0.8 cm s^{- 1}$

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