Question

One end of a horizontal rope is attached to a prong of anelectrically driven tuning fork that vibrates at 120 Hz. The otherend passes over a pulley and supports a 1.70 kg mass. The linear mass density of the rope is0.0590 kg/m.

(a) What is the speed of a transverse wave onthe rope?
v_{1.70 kg} =1 m/s

(b) What is the wavelength?
λ_{1.70 kg} =2 m

(c) How would your answers to parts (a) and (b) be changed if themass were increased to 2.80 kg?

v2.80 kg	= 3v1.70 kg
λ2.80kg	= 4λ1.70 kg

Answer 1

Concepts and reason

The concepts required to solve the given problem are speed and wavelength of a transverse wave.

First, calculate the tension in the rope by using the supported mass and acceleration due to gravity. Then determine the speed of the transverse wave on the rope by using the tension in the rope and linear mass density of the rope. The wavelength is determined by using relation of speed with frequency and wavelength.

Fundamentals

The speed v of a transverse wave on a rope is given as follows:

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

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

The speed v of a transverse wave is equal to the product of frequency f and wavelength $\lambda$of the wave.

$v = f\lambda$

(a)

The expression for speed v of a transverse wave in terms of tension T and linear density $\mu$is given as follows:

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

The tension T in the rope is equal to weight of mass m.

$T = mg$

Here, g is the acceleration due to gravity.

Substitute mg for T in the equation $v = \sqrt {\frac{T}{\mu }} .$

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

Substitute 1.70 kg for m, $9.8{\rm{ m/}}{{\rm{s}}^2}$for g, and $0.0590{\rm{ kg/m}}$for $\mu$ in equation $v = \sqrt {\frac{{mg}}{\mu }} .$

$\begin{array}{c}\\{v_{1.70{\rm{ kg}}}} = \sqrt {\frac{{\left( {1.70{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {0.0590{\rm{ kg/m}}} \right)}}} \\\\ = 16.8{\rm{ m/s}}\\\end{array}$

(b)

The speed v of a transverse wave in terms of frequency f and wavelength $\lambda$is,

$v = f\lambda$

Rearrange the above equation for $\lambda .$

$\lambda = \frac{v}{f}$

Substitute 16.8 m/s for $v$ and 120 Hz for f in the above equation.

$\begin{array}{c}\\{\lambda _{1.70{\rm{ kg}}}} = \frac{{16.8{\rm{ m/s}}}}{{120{\rm{ Hz}}}}\\\\ = 0.140{\rm{ m}}\\\end{array}$

(c.1)

The speed of a transverse wave on the rope is given as,

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

Substitute 2.80 kg for m,$9.8{\rm{ m/}}{{\rm{s}}^2}$for g, and $0.0590{\rm{ kg/m}}$for $\mu$ in equation $v = \sqrt {\frac{{mg}}{\mu }}$.

$\begin{array}{c}\\{v_{2.80{\rm{ kg}}}} = \sqrt {\frac{{\left( {2.80{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {0.0590{\rm{ kg/m}}} \right)}}} \\\\ = 21.6{\rm{ m/s}}\\\end{array}$

Divide the speed ${v_{2.80{\rm{ kg}}}}$ by speed ${v_{1.70{\rm{ kg}}}}$ and determine the change in speed.

$\begin{array}{c}\\\frac{{{v_{2.80{\rm{ kg}}}}}}{{{v_{1.70{\rm{ kg}}}}}} = \frac{{21.6{\rm{ m/s}}}}{{16.8{\rm{ m/s}}}}\\\\{v_{2.80{\rm{ kg}}}} = \left( {1.29} \right){v_{1.70{\rm{ kg}}}}\\\end{array}$

Here, ${v_{1.70{\rm{ kg}}}}$ is the speed of the wave when mass is 1.70 kg.

(c.2)

The wavelength of the wave is given as,

$\lambda = \frac{{{v_{2.80{\rm{ kg}}}}}}{f}$

Substitute 21.6 m/s for ${v_{2.80{\rm{ kg}}}}$ and 120 Hz for f in the above equation.

$\begin{array}{c}\\{\lambda _{2.80{\rm{ kg}}}} = \frac{{21.6{\rm{ m/s}}}}{{120{\rm{ Hz}}}}\\\\ = 0.180{\rm{ m}}\\\end{array}$

The change in the wavelength is determined by dividing the wavelength ${\lambda _{2.80{\rm{ kg}}}}$by wavelength ${\lambda _{1.70{\rm{ kg}}}}.$

$\begin{array}{c}\\\frac{{{\lambda _{2.80{\rm{ kg}}}}}}{{{\lambda _{1.70{\rm{ kg}}}}}} = \frac{{0.180{\rm{ m}}}}{{0.140{\rm{ m}}}}\\\\{\lambda _{2.80{\rm{ kg}}}} = \left( {1.29} \right){\lambda _{1.70{\rm{ kg}}}}\\\end{array}$

Ans: Part a

The speed of a transverse wave on the rope is 16.8 m/s.

Part b

The wavelength of the wave is 0.140 m.

Part c.1

The speed of transverse wave is increased by a factor 1.29.

Part c.2

The wavelength of transverse wave is changed by a factor 1.29.