Question

In: Physics

The wave function for a traveling wave on a taut string is (in SI units) y(x,t)...

The wave function for a traveling wave on a taut string is (in SI units)
y(x,t) = 0.375 sin (14pt - 2px + p/4)

(a) What are the speed and direction of travel of the wave?
speed _____ m/s
direction_________
(positive x-direction, positive y-direction, positive z-direction, negative x-direction, negative y-direction, negative z-direction)

(b) What is the vertical position of an element of the string at t = 0, x = 0.178 m?
________m

(c) What is the wavelength of the wave?
____________m

(d) What is the frequency of the wave?
________ Hz

(e) What is the maximum transverse speed of an element of the string?
_____ m/s

Solutions

Expert Solution

Concepts and reason

The concept used in this problem is equation of wave. The characteristics of a wave like amplitude, wavelength, speed and frequency are also used here. Use the general wave equation to answer the given questions.

Fundamentals

Equation of wave:

The general wave equation is:

y=Asin(ωtkx+ϕ)y = A\sin \left( {\omega t - kx + \phi } \right)

Here, AA is the amplitude of wave, kk is the wave number, xx is the distance covered in given time, ω\omega is the angular frequency, tt is the time taken and ϕ\phi is the phase constant.

The maximum displacement of the particles which are vibrating from its mean position is the amplitude of the wave.

“The wavelength is the distance between two successive crests or troughs.” The relation between wavelength and wave number is:

λ=2πk\lambda = \frac{{2\pi }}{k}

The speed with which a wave travels is the transverse speed of wave. The expression for speed is:

v=ωkv = \frac{\omega }{k}

The frequency of a wave is inversely proportional to the wavelength. The expression for frequency is:

f=vλf = \frac{v}{\lambda }

The relation between maximum transverse speed and angular frequency is:

vmax=Aω{v_{\max }} = A\omega

(a)

The speed of wave is:

v=ωkv = \frac{\omega }{k}

Substitute 14πrad/s14\pi {\rm{ rad/s}} for ω\omega and 2πradm12\pi {\rm{ rad}} \cdot {{\rm{m}}^{ - 1}} for kk .

v=14πrad/s2πradm1=7m/s\begin{array}{c}\\v = \frac{{14\pi {\rm{ rad/s}}}}{{2\pi {\rm{ rad}} \cdot {{\rm{m}}^{ - 1}}}}\\\\ = {\bf{7 m/s}}\\\end{array}

(b)

The given wave equation is:

y=0.375sin(14πt2πx+π/4)y = 0.375\sin \left( {14\pi t - 2\pi x + \pi /4} \right)

Substitute 0.178m0.178\;{\rm{m}} for xx and 00 for tt .

y=0.1225my = - {\bf{0}}{\bf{.1225}}\;{\bf{m}}

(c)

The wavelength is:

λ=2πk\lambda = \frac{{2\pi }}{k}

Substitute 2πradm12\pi {\rm{ rad}} \cdot {{\rm{m}}^{ - 1}} for kk .

λ=2π2πradm1=1m\begin{array}{c}\\\lambda = \frac{{2\pi }}{{2\pi {\rm{ rad}} \cdot {{\rm{m}}^{ - 1}}}}\\\\ = {\bf{1}}\;{\bf{m}}\\\end{array}

(d)

The frequency of wave is:

f=vλf = \frac{v}{\lambda }

Substitute 7m/s7\;{\rm{m/s}} for vv and 1m1\;{\rm{m}} for λ\lambda .

f=7m/s1m=7Hz\begin{array}{c}\\f = \frac{{7\;{\rm{m/s}}}}{{1\;{\rm{m}}}}\\\\ = {\bf{7}}\;{\bf{Hz}}\\\end{array}

(e)

The maximum transverse speed of wave is:

vmax=Aω{v_{\max }} = A\omega

Substitute 0.375m0.375\;{\rm{m}} for AA and 14πrad/s14\pi \;{\rm{rad/s}} for ω\omega .

vmax=(0.375m)(14πrad/s)=16.485m/s\begin{array}{c}\\{v_{\max }} = \left( {0.375\;{\rm{m}}} \right)\left( {14\pi \;{\rm{rad/s}}} \right)\\\\ = {\bf{16}}{\bf{.485}}\;{\bf{m/s}}\\\end{array}

Ans: Part a

The speed of wave is 7 m/s and in positive x-direction.

Part b

The vertical position of an element is -0.1225 m.

Part c

The wavelength is 1 m.

Part d

The frequency of wave is 7 Hz.…/.

Part e

The maximum speed of an element of spring is 16.485 m/s.…/.


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