Question

A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves up and down through a distance of 1.00 cm. The motion is continuous and is repeated regularly 120 times per second. The string has linear density 120 g/mand is kept under a tension of 90.0 N. Find the maximum value of (a) the transverse speed uand (b) the transverse component of the tensiont. (c) Show that the two maximum values calculated above occur at the same phase values for the wave. What is the transverse displacement yof the string at these phases? (d) What is the maximum rate of energy transfer along the string? (e) What is the transverse displacement ywhen this maximum transfer occurs? (f) What is the minimum rate of energy transfer along the string? (g) What is the transverse displacement ywhen this minimum transfer occurs?

Short answer

1. The maximum value of the transverse speed is 3.77 m/s .
2. The maximum value of the transverse component of the tension is 12.3 N.
3. The maximum value calculated above occurs at the same phase values for the wave, so the transverse displacement y of the string at these phases is 0.
4. The maximum rate of energy transfer along the string is 46.4 W.
5. The transverse displacement y when this maximum transfer occurs is 0.
6. The minimum rate of energy transfer along the string is when the transverse component of tension and speed is zero.
7. The transverse displacement y when this minimum transfer occurs is$\pm 0.50\mathrm{cm}$ .

Step-by-step solution

The given data

1. The movement of the bar is, d = 1.00 cm .
2. Frequency is, f = 120 Hz.
3. Linear mass density is, $\mu =120\mathrm{g}/\mathrm{m}$.
4. Tension is, T = 90.0 N .

Understanding the concept of the wave equation

We use the concept of wave motion. Using the equation of the wave, we can find its derivative to calculate speed. Using the equation of speed related to tension and linear mass density, we find the component of tension. We can find the values of displacement and energy transfer by finding the phase values. We can find energy transfer that is the power from work done and displacement. Finally, we find displacement at minimum power.

Formulae:

The general expression of the wave,

$y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)$ (i)

The velocity of the wave,

$v=\sqrt{\frac{T}{\mu }}$ (ii)

The tangent angle,

$\tan \theta =\frac{dy}{dx}$ (iii)

The maximum rate of energy transfer in a body,

$P=\frac{W}{t}$ (iv)

Here,$y_m$ is the amplitude of the wave, k is the wave number, $\omega$is the angular frequency of the wave, and W is the work done.

a) Calculation of maximum value of transverse speed

The displacement equation is given by equation (i) is,

$y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)$

The velocity of the wave is given by the derivative of the displacement as:

$v\left(x,t\right)=\frac{dy}{dt}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} v\left(x,t\right)=\frac{d}{dt}{y}_m\sin \left(kx-\omega t\right) \\ =-y_m\omega \cos \left(kx-\omega t\right) \end{gathered}$$

For the maximum value of speed, the term role="math" localid="1661166121785" $\cos \left(kx-\omega t\right)$should be maximum, and it is maximum at $\left(kx-\omega t\right)$= 0 , we get,

$v_{max}=-y_m\omega$ (a)

Hereis the amplitude; we can find it from the given movement of the bar as:

$$\begin{gathered} y_m=\frac{d}{2} \\ =\frac{1.00}{2} \\ =0.50\mathrm{cm} \\ =5\times {10}^{-3}\mathrm{m} \end{gathered}$$

The angular frequency of the wave can be calculated as,

$\omega =2\pi f$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \omega =2\left(3.14\right)\left(120\right) \\ =754\mathrm{rad}/\mathrm{s} \end{gathered}$$

The maximum speed of the wave using equation (a), we get,

$$\begin{gathered} v_{max}=\left(5\times {10}^{-3}\times 754\right) \\ =3.77\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the maximum transverse speed is 3.77 m/s .

b) Calculation of maximum transverse component of tension

We can find the derivative of the displacement equation with respect to x, and we get,

$$\begin{gathered} \frac{dy}{dx}=\frac{d\left(y_m\sin \left(kx-\omega t\right)\right)}{dx} \\ =y_{m}k\cos \left(kx-\omega t\right) \end{gathered}$$

Hence, from equation (iii) and the above value of the derivative, we get

$\tan \theta =y_{m}k\cos \left(kx-\omega t\right)$

For the maximum value of $\tan \theta$term $\cos \left(kx-\omega t\right)$should be maximum, and it is maximum at$\left(kx-\omega t\right)=0$, we get,

$\tan \theta =y_{m}k$ (v)

Here we need to find the wave number k, that is,

$k=\frac{\omega }{v}$ (b)

Substitute the values in the equation (ii); we get the velocity of the wave as:

$$\begin{gathered} v=\sqrt{\frac{90.0}{0.120}} \\ =27.4\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute the values in equation (b); we can calculate the wave number as:

$\begin{array}{l}k=\frac{754}{27.4}\\ =27.5m^{-1}\end{array}$

Substitute the values in the equation (v); we can calculate the angle as:

$\begin{array}{l}\tan \theta =5=5\times {10}^{-3}\times 27.5\\ =0.138\\ \theta ={\tan }^{-1}\left(0.138\right)\\ =7.{83}^0\end{array}$

So the maximum transverse component of tension can be calculated as;

$$\begin{gathered} T_{Trans}=T\sin \theta \\ =90.0\times \sin (7.83°) \\ =12.3\mathrm{N} \end{gathered}$$

Hence, the value of transverse speed is 12.3 N .

 Step 5: c) Calculation of the transverse displacement

The transverse component of tension will pull the string left; the equation for the can be written as,

$-T\left(\frac{dy}{dx}\right)=-Tk{y}_m\cos (kx-\omega t)$

It reaches a maximum value when$\cos (kx-\omega t)=-1$.

The two quantities reach their maximum values at the same value of phase. When the term$\cos (kx-\omega t)=-1$the value of the term$\sin (kx-\omega t)$will be zero, the displacement of the string y = 0 .

So, from the equation of displacement,

$y(x,t)=y_m\sin \left(kx-\omega t\right)$

At$\left(kx-\omega t\right)$= 0 the value of $\sin \left(kx-\omega t\right)$is zero; hence, at this value, we get transverse displacement as 0.

d) Calculation of maximum energy rate of transfer

The tension does work when a point moves by small displacement$∆y$in time$∆t$, then the work done is,

$dW=T_{trans}\times ∆y$

From equation (iv), we can write the expression for energy rate transfer as,

$P=\frac{T_{trans}∆y}{∆t}$

We know that the velocity of the wave can be written as,

$\frac{∆y}{∆t}=v$

Hence, the energy transfer is given as:

$P=T_{trans}v$

For maximum power, the tension component and the speed should be maximum; we can calculate the maximum value of energy rate as:

$$\begin{gathered} P=12.3\times 3.77 \\ =46.4W \end{gathered}$$

Hence, the value of the rate of energy is 46.4 W .

e) Calculation of transverse displacement when maximum energy transfer occurs

From the above calculations, we know that maximum transfer occurs at maximum values of transverse displacement and speed, so at that value y(x,t) = 0 .

Thus, the transverse displacement y when this maximum transfer occurs is 0.

f) Calculation of minimum rate of energy

When the transverse component of tension and speed are zero, we get the minimum rate of energy transfer along the string.

Thus, the minimum rate of energy transfer along the string, when the transverse component of tension and speed are zero, is zero.

g) Calculation of transverse displacement when minimum energy transfer occurs

When power transfer is minimum, that is zero, the values of$\cos \left(kx-\omega t\right)=0$and $\sin \left(kx-\omega t\right)=\pm 1$ .

In this case, the string displacement can be calculated as,

$$\begin{gathered} y\left(x,t\right)=\pm y_m \\ =\pm 0.50\mathrm{cm} \end{gathered}$$

Hence, the value of transverse displacement is$\pm 0.50\mathrm{cm}$