Question

In Problem 55 (Fig. 11–55), the length l of the string may be adjusted by moving the pulley. If the hanging mass m is fixed at 0.080 kg, how many different standing wave patterns may be achieved by varying l between 10 cm and 1.5 m?

Short answer

Three standing wave patterns are achieved between \(10{\rm{ cm}}\) and \(1.5{\rm{ m}}\).

Step-by-step solution

Wavelength of the string

The relationship between the frequency and the string's tension is used in this problem. The length of the string is half the wavelength.

Given data

The hanging mass is \(m = 0.080{\rm{ kg}}\).

The frequency is \(f = 60{\rm{ Hz}}\).

The string’s mass per unit length is \(\mu = 3.5 \times {\rm{1}}{{\rm{0}}^{ - 4}}{\rm{ kg/m}}\).

Calculation of the wavelength

The wavelength of the standing wave is calculated as,

\(\begin{aligned}{l}\lambda = \frac{1}{f}\sqrt {\frac{F}{\mu }} \\\lambda = \frac{1}{f}\sqrt {\frac{{mg}}{\mu }} \end{aligned}\)

Here, F is the tension in the string.

Substitute the values in the above relation.

\(\begin{aligned}{c}\lambda &= \frac{1}{{60{\rm{ Hz}}}}\sqrt {\frac{{\left( {0.080{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {3.5 \times {{10}^{ - 4}}{\rm{ kg/m}}} \right)}}} \\ &= 0.788{\rm{ m}}\end{aligned}\)

Calculation of the number of patterns in the given length

For the first harmonic, the length of the string is,

\(\begin{aligned}{c}l &= \frac{\lambda }{2}\\ &= \left( {\frac{{0.788{\rm{ m}}}}{2}} \right)\\ &= 0.394{\rm{ m}}\end{aligned}\)

For the second harmonic, the length of the string is,

\(\begin{aligned}{c}l &= \lambda \\ &= 0.788{\rm{ m}}\end{aligned}\)

For the third harmonic, the length of the string is,

\(\begin{aligned}{c}l &= \frac{{3\lambda }}{2}\\ &= \frac{{3 \times 0.788{\rm{ m}}}}{2}\\ &= 1.182{\rm{ m}}\end{aligned}\)

Thus, between \(10{\rm{ cm}}\) and \(1.5{\rm{ m}}\) three standing wave patterns can achieve.