Respuesta :
The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
[tex]y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])[/tex]
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
[tex]y(x,t)=0.750 cm[/tex]
and therefore this value corresponds to the amplitude of the wave.
In our problem, the equation of the wave is
[tex]y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])[/tex]
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
[tex]y(x,t)=0.750 cm[/tex]
and therefore this value corresponds to the amplitude of the wave.
The amplitude of the given transverse wave in the rope is [tex]\boxed{0.750\,{\text{cm}}}[/tex].
Further Explanation:
The given expression of the wave developed in the rope is.
[tex]y\left( {x,t} \right) = \left( {0.750\,{\text{cm}}} \right)\cos \left( {\pi \left[ {\left( {0.400\,{\text{cm}} - {\text{1}}} \right)x + \left( {250\,{\text{s}} - 1} \right)t} \right]} \right)[/tex]
The standard equation of the wave produced in a rope is given as.
[tex]\boxed{y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)}[/tex]
Here, [tex]y[/tex] is the position of the wave, [tex]A[/tex] is the amplitude of the wave, [tex]k[/tex] is the wave number and [tex]\omega[/tex] is the angular frequency of the wave.
On comparing the above equation of the wave with the standard equation of the wave produced in a rope, the amplitude of the wave can be obtained as follows.
[tex]A = 0.750\,{\text{cm}}[/tex]
Thus, the amplitude of the given transverse wave in the rope is [tex]\boxed{0.750\,{\text{cm}}}[/tex].
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Answer Details:
Grade: High School
Chapter: Waves
Subject: Physics
Keywords: Transverse, wave, amplitude, angular frequency, stationary wave, rope, velocity, position, wave number.
