A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). Part A Find the amplitude….
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Question “A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). Part A Find the amplitude….”
A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]).
Part A Find the amplitude.
Part B Find the period
Part C Find the frequency
Part D Find the wavelength
Part E Find the speed of propagation.
Part F Is the wave traveling in the +x- or − x-direction?
Answer
This problem can be solved using the wave-equations of a wave traveling in both positive and negative direction in terms of amplitude and wavelength, frequency, time, and position.
Write the general wave equation, describing amplitude, wavelength frequency, frequency, time, and position.
Next, you can compare the general wave equation with the given wave equation to determine the amplitude and period, frequency, wavelengths, speed of propagation, direction of propagation, as well as the frequency and frequency of the wave.
Here is the wave equation for wave traveling in positive xdirection:
y\left( {x,t} \right) = A\cos \left( {kx - \omega t} \right)Here, y represents the position at which the wave is at y-coordinate at time t. \omega is the wave-vector for the wave.
Here is the wave equation for wave traveling in negative xdirection:
y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)Here is the relation between wavelength and wavevector:
k = \frac{{2\pi }}{\lambda }Here, \lambda
Here is the relationship between frequency and angular frequency:
\omega = 2\pi ff is the wavelength of the wave.
These are the parameters that describe the speed of propagation of waves in terms of frequency and wavevector.
v = \frac{\omega }{k}v indicates the velocity of the wave.
Here is the relationship between frequency and time-period for the wave:
T = \frac{1}{f}T refers to the time period of the wave.
(Part A)
Here is the wave-equation for wave travelling in negative x-direction:
y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)Here is the wave-equation for problem:
y\left( {x,t} \right) = \left( {0.750{\rm{ cm}}} \right)\cos \left( {\pi \left( {0.400{\rm{ c}}{{\rm{m}}^{ - 1}}} \right)x + \pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)t} \right)Compare the wave-equation for wave travelling in negative x direction and wave-equation problem to determine the wave’s amplitude.
The wave’s amplitude is therefore 0.750 cm.
Part B
Here is the wave-equation for wave travelling in negative x-direction:
y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)Here is the wave-equation for problem:
y\left( {x,t} \right) = \left( {0.750{\rm{ cm}}} \right)\cos \left( {\pi \left( {0.400{\rm{ c}}{{\rm{m}}^{ - 1}}} \right)x + \pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)t} \right)Compare the wave-equation for wave travelling in negative x direction and wave-equation problem to determine the angular frequency.
The angular frequency for the wave is therefore $math_tag_12
Here is the relationship between the angular frequency of the wave and its time-period:
T = \frac{{2\pi }}{\omega }Substitute $math_tag_12
\begin{array}{c}\\T = \frac{{2\pi }}{{\left( {250\pi {\rm{ }}{{\rm{s}}^{ - 1}}} \right)}}\\\\ = 0.008{\rm{ s}}\\\end{array}The time period of the wave is therefore 0.008 seconds.
Part C
Here is the relationship between frequency and time period:
f = \frac{1}{T}Substitute T for 0.008 s in the equation above.
\begin{array}{c}\\f = \frac{1}{{\left( {0.008{\rm{ s}}} \right)}}\\\\ = \left( {125{\rm{ }}{{\rm{s}}^{ - 1}}} \right)\left( {\frac{{{\rm{Hz}}}}{{1{\rm{ }}{{\rm{s}}^{ - 1}}}}} \right)\\\\ = 125{\rm{ Hz}}\\\end{array}The frequency of the wave therefore is 125 Hz.
Part D
Here is the wave-equation for wave travelling in negative x-direction:
y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)Here is the wave-equation for problem:
y\left( {x,t} \right) = \left( {0.750{\rm{ cm}}} \right)\cos \left( {\pi \left( {0.400{\rm{ c}}{{\rm{m}}^{ - 1}}} \right)x + \pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)t} \right)Compare the wave-equation for wave travelling in negative x direction and wave-equation with problem to determine the wavevector.
The wavevector is therefore $math_tag_20
Here is the relationship between wavelength and wavevector:
\lambda = \frac{{2\pi }}{k}Substitute $math_tag_20
\begin{array}{c}\\\lambda = \frac{{2\pi }}{{\left( {0.400\pi {\rm{ c}}{{\rm{m}}^{ - 1}}} \right)}}\\\\ = 5{\rm{ cm}}\\\\ = 5{\rm{ cm}}\\\end{array}The wavelength of the wave therefore is 5 cm
Part E
Here is the relation between wavevector frequency and angular frequency:
v = \frac{\omega }{k}Substitute $math_tag_20
\begin{array}{c}\\v = \frac{\omega }{k}\\\\ = \frac{{250\pi {\rm{ }}{{\rm{s}}^{ - 1}}}}{{0.400\pi {\rm{ c}}{{\rm{m}}^{01}}}}\\\\ = 625{\rm{ cm/s}}\\\end{array}Accordingly, the speed of the wave at $math_tag_27 is
Part F
Here is the wave equation for wave traveling in positive xdirection:
y\left( {x,t} \right) = A\cos \left( {kx - \omega t} \right)Here is the wave equation for wave traveling in negative xdirection:
y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)Here is the wave-equation for problem:
y\left( {x,t} \right) = \left( {0.750{\rm{ cm}}} \right)\cos \left( {\pi \left( {0.400{\rm{ c}}{{\rm{m}}^{ - 1}}} \right)x + \pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)t} \right)Compare the wave equation given with those travelling in negative x-directions. The wave is traveling in a negative x direction.
The direction of propagation is therefore in the negative x direction.
Ans Part A
The wave’s amplitude is 0.750 cm.
Part B
The time period of the wave is 0.008 seconds.
Part C
The frequency of the wave is at 125 Hz.
Part D
The wave’s wavelength is 5 cm.
Part E
The speed of the wave {\bf{625 cm/s}}Part A
Wave propagation direction is in the negative and the positive directions.
Conclusion
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