Q19E - University Physics with Modern Physics

Question

A thin, 75.0cm wire has a mass of 16.5g. One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire. (a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 cm makes 625 vibrations per second? (b) How fast would this wave travel?

Step-by-Step Solution

Answer

Hence, the tension in the screw must be adjusted to 9.53N so that transverse wave of wavelength 3.33 cm makes 625 vibrations per second. This wave will travel at a speed of 20.8m/s.

Step 1: Determination of the formula of Mechanical Waves

To find the tension in wire we can use the dynamic relationship,

$v=\sqrt{\frac{T}{\mu }}$--(1)

To find the wave speed from the kinematic relationship we can use, 

$v=f\lambda$ 

Step 2: Calculation of the tension in the screw

The rope's linear mass density is 
 

$$\begin{gathered} \mu =\frac{m_{rope}}{L} \\ =\frac{16.5g}{75.0cm} \\ =0.22g/cm \end{gathered}$$
 
 From equation (1):

$v=\sqrt{\frac{T}{\mu }}$

 But we know that 

$v=f \lambda$

Therefore,

$f \lambda =\sqrt{\frac{T}{\mu }}$ 
 
 Solve for T and substituting the known values off and lambda:
 
$$\begin{gathered} T=\mu f^2 {\lambda }^2 \\ =\left(0.22g/cm\right)\left(625vib/s^2\right){\left(3.33cm\right)}^2 \\ =9.53\times {10}^5 g.cm/s^2 \end{gathered}$$ 

 

Since g-cm/s²=10, then the tension (in newtons) is:

 

 $T=9.53N$.

Step 3: Calculation of speed of the wave

The wave speed is simply given by 

$$\begin{gathered} v=f \lambda \\ =\left(625vib/s\right)(3.33\times \frac{1m}{100cm}) \\ =20.8m/s \end{gathered}$$

Therefore, v = 20.8 m/s.
 

Hence, the tension in the screw must be adjusted to 9.53N so that transverse wave of wavelength 3.33 cm makes 625 vibrations per second. This wave will travel at a speed of 20.8m/s.