91Ó°ÊÓ

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

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, $V=\sqrt{\frac{T}{\mathrm{μ}}}$ …(¾±)

The linear density of a material, $\mathrm{μ}=\frac{∆m}{∆l}$ …(¾±¾±)

The tangential acceleration of a body, $a=\frac{v{\text{'}}^2}{R}$ …(¾±¾±¾±)

The force according to Newton’s second law, F = ma …(¾±±¹)

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

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

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

$∆\mathrm{m}=\mathrm{μ}∆l$

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:

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

But, speed of the transverse waves on the string is

$v\text{'}=\sqrt{\frac{T}{\mathrm{μ}}}$

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 $5.00\mathrm{cm}/\mathrm{s}$.