Answer
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Hint: Standing wave is created when two oppositely travelling waves(with the same frequency) interfere with each other. Peaks of standing waves don’t move specially but oscillates w.r.t time. So, for a particular position maximum amplitude remains always the same.
Complete step by step answer:
Suppose, there are two oppositely travelling waves with the equations \[{y_1} = A\sin (kx - \omega t)\] and ${y_2} = A\sin (kx + \omega t)$. Now, the resulting standing wave equation created by them is $y = 2A\sin (kx)\cos (\omega t)$ (1)
where, amplitude at any point x is given by: $2A\sin (kx)$.
Here given: Standing wave equation is given as: \[y = 4\sin \left( {\dfrac{{\pi x}}{{12}}} \right)\cos \left( {52\pi t} \right)\] To find: Amplitude of the particle at x=2 cm.
Step 1
In the given equation of standing wave, put x=2 to get the equation as:
\[
y = 4\sin \left( {\dfrac{{\pi \times 2}}{{12}}} \right)\cos \left( {52\pi t} \right) \\
= 4 \times \dfrac{1}{2} \times \cos \left( {52\pi t} \right) = 2\cos \left( {52\pi t} \right) \\
\] (2)
Now, comparing the standing wave equation of eq.(2) with eq.(1) we get the amplitude as 2cm.
Correct answer:
Amplitude of the particle at x=2 cm is (c) 2cm.
Note: For a standing wave equation there is no sine or cosine term containing both variables x and t together. This implies that these waves don’t propagate in any direction like a normal wave. Now, while trying to find the amplitude of the wave a student might get confused as there are two sinusoidal terms ( both sine and cosine) in the equation. Always remember that amplitude of a wave can never be time dependent (the cosine term in eq.(1)), because it is concerned with the time dimension and not with the amplitude. Hence, the remaining part is the amplitude and that has oscillatory behavior specially.
Complete step by step answer:
Suppose, there are two oppositely travelling waves with the equations \[{y_1} = A\sin (kx - \omega t)\] and ${y_2} = A\sin (kx + \omega t)$. Now, the resulting standing wave equation created by them is $y = 2A\sin (kx)\cos (\omega t)$ (1)
where, amplitude at any point x is given by: $2A\sin (kx)$.
Here given: Standing wave equation is given as: \[y = 4\sin \left( {\dfrac{{\pi x}}{{12}}} \right)\cos \left( {52\pi t} \right)\] To find: Amplitude of the particle at x=2 cm.
Step 1
In the given equation of standing wave, put x=2 to get the equation as:
\[
y = 4\sin \left( {\dfrac{{\pi \times 2}}{{12}}} \right)\cos \left( {52\pi t} \right) \\
= 4 \times \dfrac{1}{2} \times \cos \left( {52\pi t} \right) = 2\cos \left( {52\pi t} \right) \\
\] (2)
Now, comparing the standing wave equation of eq.(2) with eq.(1) we get the amplitude as 2cm.
Correct answer:
Amplitude of the particle at x=2 cm is (c) 2cm.
Note: For a standing wave equation there is no sine or cosine term containing both variables x and t together. This implies that these waves don’t propagate in any direction like a normal wave. Now, while trying to find the amplitude of the wave a student might get confused as there are two sinusoidal terms ( both sine and cosine) in the equation. Always remember that amplitude of a wave can never be time dependent (the cosine term in eq.(1)), because it is concerned with the time dimension and not with the amplitude. Hence, the remaining part is the amplitude and that has oscillatory behavior specially.
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