Question

In Fig. 16-42, a string, tied to a sinusoidal oscillator at Pand running over a support at Q, is stretched by a block of mass m.The separation between Pand Qis 1.20 m, and the frequency fof the oscillator is fixed at 120 Hz. The amplitude of the motion atPis small enough for that point to be considered a node. A node also exists at Q. A standing wave appears when the mass of the hanging block is 286.1 gor 447.0 g, but not for any intermediate mass. What is the linear density of the string?

Short answer

Linear density of the string is 0.845 g/m

Step-by-step solution

Given data

The separation between P and Q, L = 1.2 m

The frequency of the oscillator, f = 120 Hz

The mass of the hanging block, $m_1=286.1g$or$m_2=447.0g$

Understanding the concept of frequency and nodes

The mass of the block is given. We know the formula for the frequency of the wave. Using this information, we find the nodes produced on the string. Using the nodes and the given information we find the linear density of the string.

Formula:

The frequency of nth harmonic oscillation, $f=\frac{n}{2L}\left(\sqrt{\frac{T}{\mu }}\right)..............\left(1\right)$

Calculation of linear density

We have,

From the Free body diagram,

T = mg

Hence, substituting the value of tension in equation (1), we get:

$$\begin{gathered} f=\frac{n}{2L}\left(\sqrt{\frac{mg}{\mu }}\right) \\ \frac{2Lf}{n}=\sqrt{\frac{mg}{\mu }} \\ {\left(\frac{2Lf}{n}\right)}^2=\frac{mg}{\mu } \\ m=\frac{\mu }{g}\times {\left(\frac{2Lf}{n}\right)}^2............\left(2\right) \end{gathered}$$

Here$\mu ,g,L$are f constants,

So,

$m\propto \frac{1}{n^2}$

The mass of the hanging block is either 286.1 g or 447 g

So,

We assume that

447 g corresponds to and 286.1 g corresponds to ( n + 1 ) , hence

$$\begin{gathered} 447\propto \frac{1}{n^2} \\ 286.1\propto \frac{1}{{\left(n+1\right)}^2} \end{gathered}$$

We take the ratio of the above equations; we get the value of n as:

$$\begin{gathered} \frac{447}{286.1}=\frac{{\displaystyle \frac{1}{n^2}}}{{\displaystyle \frac{1}{{\left(n+1\right)}^2}}} \\ 1.56=\frac{{\left(n+1\right)}^2}{n^2} \\ 1.56n^2=n^2+2n+1 \\ 0.56n^2-2n-1=0 \end{gathered}$$

Solving the quadratic equation, we get the two values for n as,

n = 4 and n = -0.44

As the negative values for n are not acceptable, so

n = 4,for mass m = 447 g or 0.447 kg

Hence, substituting the given values and the value of n in equation (2), we get the linear density as given:

role="math" localid="1661157160876" $$\begin{gathered} \mu =0.447\times 9.8\times {\left(\frac{4}{2\times 1.2\times 20}\right)}^2 \\ =8.45\times \frac{{10}^{-4}kg}{m} \\ =0.845g/m \end{gathered}$$

Hence, the value of linear density is$0.845g/m$