Question

A wire with mass 45.0 g is stretched so that its ends are tied down at points a distance 81.0 cm apart. The wire vibrates in its fundamental mode with frequency 62.0 Hz and with an amplitude at the antinodes of 0.310 cm.

a) What is the speed of propagation of transverse waves in the wire?

v = ______ m/s

b) Compute the tension in the wire.

F = _____ N

c) Find the magnitude of the maximum transverse velocity of particles in the wire.

= _____ m/s

d) Find the magnitude of the maximum acceleration of particles in the wire.

= m/s2

The answers i have are these:-

a) 96.0

b) 461

d) 426

 

Answer 1

Concepts and reason

The concept required to solve this problem is transverse wave equations.

Initially use the equation of velocity of wave and substitute wavelength of fundamental in terms of length of wire to calculate the velocity.

Later, rearrange the velocity equation in terms of tension and calculate the tension. Then, use the equation of maximum transverse velocity and substitute angular speed in the expression and calculate the maximum transverse velocity.

Finally use the equation of maximum transverse acceleration and substitute angular speed in the expression and calculate the maximum transverse acceleration.

Fundamentals

The velocity v of a transverse wave is given as follows:

$v = \lambda f$

Here, $\lambda$ is the wavelength, and $f$ is the frequency.

The wavelength for a fundamental mode is equal to the 2 times that of length l of the string.

$\lambda = 2l$

The velocity in terms of Tension force is given as follows:

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

Here, $F$ is the tension in the string, and $\mu$ is the mass per unit length of the string.

The mass per unit length of wire is the ratio of mass and length of the wire.

$\mu = \frac{m}{l}$

Here, m is the mass of the wire, and l is the length of wire.

The magnitude of maximum transverse velocity ${v_{\max }}$of particles is,

${v_{\max }} = A\omega$

Here, $A$ is the amplitude, and $\omega$ is the angular speed.

The magnitude of maximum acceleration ${a_{\max }}$of particles is,

${a_{\max }} = A{\omega ^2}$

Here, $A$ is the amplitude, and $\omega$ is the angular speed.

The angular speed of the wave is given as,

$\omega = 2\pi f$

Here, $f$ is the frequency.

(a)

The speed of propagation of transverse wave in the wire is,

$v = \lambda f$

Substitute $2l$ for $\lambda$ in the equation $v = \lambda f.$

$v = 2lf$

Substitute $81.0{\rm{ cm}}$ for $l,$ and $62.0{\rm{ Hz}}$ for $f,$ in the equation $v = 2lf.$ $\begin{array}{c}\\v = 2\left( {81.0{\rm{ cm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\left( {62.0{\rm{ Hz}}} \right)\\\\ = 100.4\,{\rm{m/s}}\\\end{array}$

(b)

The tension the wire is given as follows:

$F = \left( {\frac{m}{l}} \right){v^2}$

Substitute $45.0{\rm{ g}}$ for m, 81.0 cm for l, and 100.44 m/s for v in the equation $F = \left( {\frac{m}{l}} \right){v^2}$ and calculate the tension force F in the wire.

$\begin{array}{c}\\F = \left( {\frac{{45.0{\rm{ g}}}}{{81.0{\rm{ cm}}}}} \right){\left( {100.44{\rm{ m/s}}} \right)^2}\\\\ = \left( {45.0{\rm{ g}}\left( {\frac{{1\,{\rm{kg}}}}{{1000\,{\rm{g}}}}} \right)} \right)\left( {\frac{1}{{81.0{\rm{ cm}}}}\left( {\frac{{100{\rm{ cm}}}}{{1{\rm{ m}}}}} \right)} \right){\left( {100.44\,{\rm{m/s}}} \right)^2}\\\\ = 560.5\,{\rm{N}}\\\end{array}$

(c)

The maximum velocity ${v_{\max }}$of transverse particles in the wire is,

$\begin{array}{c}\\{v_{\max }} = A\left( {2\pi f} \right)\\\\ = 2\pi Af\\\end{array}$

Substitute $0.310{\rm{ cm}}$ for $A$, and $62.0{\rm{ Hz}}$ for $f$ in the equation ${v_{\max }} = 2\pi Af$ to calculate the magnitude of the maximum transverse velocity of particles in the wire.

$\begin{array}{c}\\{v_{\max }} = 2\pi \left( {0.31\,{\rm{cm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)\left( {62.0\,{\rm{Hz}}} \right)\\\\ = 1.21\,{\rm{m/s}}\\\end{array}$

(d)

The maximum acceleration of transverse particles in the wire is,

${a_{\max }} = 4{\pi ^2}{f^2}A$

Substitute $0.310{\rm{ cm}}$ for $A$, and $62.0{\rm{ Hz}}$ for $f$ in the equation ${a_{\max }} = A{\left( {2\pi f} \right)^2}$ to calculate the magnitude of the maximum transverse acceleration of particles in the wire.

$\begin{array}{c}\\{a_{\max }} = 4{\pi ^2}{\left( {62.0\,{\rm{Hz}}} \right)^2}\left( {0.31\,{\rm{cm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)\\\\ = 470.4\,{\rm{m/}}{{\rm{s}}^2}\\\end{array}$

Ans: Part a

The speed of propagation of transverse waves in the wire is $100.4\,{\rm{m/s}}{\rm{.}}$

Part b

The tension in the wire is 560.5 N.

Part c

The magnitude of the maximum transverse velocity of particles in the wire is $1.21{\rm{ m/s}}{\rm{.}}$

Part d

The magnitude of the maximum transverse acceleration of particles in the wire is$470.4{\rm{ m/}}{{\rm{s}}^2}.$