Question

In Fig. 16-50, a circular loop of string is set spinning about the center point in a place with negligible gravity. The radius is 4.00 cmand the tangential speed of a string segment is 5.00cm/s. The string is plucked. At what speed do transverse waves move along the string? (Hint:Apply Newton’s second law to a small, but finite, section of the string.)

Short answer

The speed at which the transverse waves move along the string is5.00cm/s .

Step-by-step solution

Identification of given data

1. Radius of the circular loop is R = 4.00 cm.
2. Tangential speed of the string segment is v' = 5.00 cm/s.

Significance of Newton’s second law

According to Newton's second law, an object's acceleration is determined by its mass and the net force that is acting on it. The acceleration of the body is inversely related to its mass and directly proportional to the net force applied on it.

We can find the expression for the tangential speed of the section of the string by applying Newton’s law to the small section of the string. Comparing it with the expression for the speed of the transverse waves moving along the string, we can find its value.

Formula:

The transverse speed of a body, $V=\sqrt{\frac{T}{\mu }}$ …(i)

The linear density of a material, $\mathrm{\mu }=\frac{∆m}{∆l}$ …(ii)

The tangential acceleration of a body, $a=\frac{v{\text{'}}^2}{R}$ …(iii)

The force according to Newton’s second law, F = ma …(iv)

Determining the speed of transverse wave

$5.00\mathrm{cm}/\mathrm{s}$Applying Newton’s law to the small section of the string, we get

$$\begin{gathered} \begin{array}{l}F=2\tau \sin \theta \\ =\tau \left(2\theta \right)\end{array}(\because \mathrm{\theta }\mathrm{is}\mathrm{very}\mathrm{very}\mathrm{small}) \\ =\frac{\tau ∆l}{R}.....................\left(b\right)\left(\because angl\mathrm{e}\mathrm{of}\mathrm{an}\mathrm{arc},\theta =\frac{∆l}{R}\right) \end{gathered}$$

From equation (ii), we can get the mass of the string as:

$∆\mathrm{m}=\mu ∆l$

Let v’ be the tangential speed of the section of the string then substituting the values of equations (a) and equation (iii), we get the force on the string as:

$$\begin{gathered} F=\mu ∆l\frac{v{\text{'}}^2}{R} \\ \mu ∆l\frac{v{\text{'}}^2}{R}=\frac{\tau ∆l}{R}\left(\because \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{force}\mathrm{from}\mathrm{equation}\left(\mathrm{b}\right)\right) \\ v\text{'}=\sqrt{\frac{T}{\mu }} \end{gathered}$$

But, speed of the transverse waves on the string is

$v\text{'}=\sqrt{\frac{T}{\mu }}$

Hence,

$$\begin{gathered} v\text{'}=v \\ =5.00\mathrm{cm}/\mathrm{s} \end{gathered}$$

Therefore, the speed at which transverse waves move along the string is $5.00\mathrm{cm}/\mathrm{s}$.