In Figure 16-36 (a), string 1 has a linear density of 3.00 g/m, and string 2 has a linear density of 5.00 g/m. They are under tension due to the hanging block of mass M = 500 g. (a)Calculate the wave speed on string 1 and (b) Calculate the wave speed on string 2. (Hint:When a string loops halfway around a pulley, it pulls on the pulley with a net force that is twice the tension in the string.) Next the block is divided into two blocks (with ${\mathit{M}}_{\mathbf{1}}\mathbf{+}{\mathit{M}}_{\mathbf{2}}\mathbf{=}\mathit{M}$) and the apparatus is rearranged as shown in Figure (b). (c) Find M1and (d) Find M2such that the wave speeds in the two strings are equal.

• Linear density of string 1,${\mu }_1=3g/m=0.003kg/m$
• Linear density of string 2, ${\mu }_2=5g/m=0.005kg/m$
• Mass of the block, M = 500 g = 0.5 kg

If we apply the tension force$\left(T\right)$ of the string, it is equally transmitted all along the

Formula:

Wave speed of the wave, $v=\sqrt{\left(\frac{T}{\mu }\right)}$ (i)

Here, T is the tension in the string and$\mu$ is the linear density of the string.

The weight of the block is equally shared by two strings, hence

$$\begin{gathered} 2T=Mg \\ T=\frac{Mg}{2}..........................\left(1\right) \end{gathered}$$

Using equation 1 in equation (i), we get the speed as:

$$\begin{gathered} v=\sqrt{\frac{Mg/2}{\mu }} \\ =\sqrt{\frac{Mg}{2\mu }}....................\left(2\right) \end{gathered}$$

To findthewave speed of spring 1, using equation 2 and the given values, we get

$$\begin{gathered} v_1=\sqrt{\frac{Mg}{2{\mu }_1}} \\ =\sqrt{\frac{0.5kg\times 9.8m/s^2}{2\times 0.00kg/m}} \\ =28.57m/s \\ \approx 28.6m/s \end{gathered}$$

Hence, the wave speed of string 1 is 28.6 m/s

To find wave speed of spring 2, using equation 2 and the given values, we get

$$\begin{gathered} v_2=\sqrt{\frac{Mg}{2{\mu }_2}} \\ =\sqrt{\frac{0.5kg\times 9.8m/s^2}{2\times 0.005kg/m}} \\ =22.13m/s \\ \approx 22.1m/s \end{gathered}$$

Hence, the value of the wave speed of string 2 is 22.1 m/s .

In the second case, the block is divided in two unequal parts but the wave speeds of the two strings are equal. Applying equation 2 for the two blocks as wave speed is equal, we get

$\sqrt{\frac{M_{1}g}{2{\mu }_1}}=\sqrt{\frac{M_{2}g}{2{\mu }_2}}$

Simplifying the equation we get,

$\frac{M_1}{{\mu }_1}=\frac{M_2}{{\mu }_2}.....................\left(3\right)$

Total mass of the block (M) = mass of the first block+ mass of the second block

$$\begin{gathered} \left(m_2\right) \\ M=M_1+M_2 \\ {M}_2=M-M_2....................\left(a\right) \end{gathered}$$

Using this value of mass in equation 3, we get

$$\begin{gathered} \frac{M_1}{{\mu }_1}=\frac{M-M_1}{{\mu }_2} \\ {M}_1=\frac{{\mu }_{1}M}{{\mu }_1+{\mu }_2} \\ =\frac{0.003kg/m\times 0.5kg}{0.003kg/m+0.005kg/m} \\ =0.1875kg \\ \approx 0.188kg \end{gathered}$$

Hence, the mass of first block is 0.1875 kg

The mass of the second block can be calculated using equation (a) as:

$$\begin{gathered} M_2=0.5kg-0.1875kg \\ =0.313kg \end{gathered}$$

Hence, mass of the second block is 0.313 kg