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 part complete find the amplitude.

Answer 1

The amplitude of a wave corresponds to its maximum oscillation of the wave itself. 
In our problem, the equation of the wave is
$y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])$
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
$y(x,t)=0.750 cm$
and therefore this value corresponds to the amplitude of the wave.

Answer 2

The amplitude of the given transverse wave in the rope is $\boxed{0.750\,{\text{cm}}}$.

Further Explanation:

The given expression of the wave developed in the rope is.

 $y\left( {x,t} \right) = \left( {0.750\,{\text{cm}}} \right)\cos \left( {\pi \left[ {\left( {0.400\,{\text{cm}} - {\text{1}}} \right)x + \left( {250\,{\text{s}} - 1} \right)t} \right]} \right)$

The standard equation of the wave produced in a rope is given as.

$\boxed{y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)}$

Here, $y$ is the position of the wave, $A$ is the amplitude of the wave, $k$ is the wave number and $\omega$ 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.

 $A = 0.750\,{\text{cm}}$

Thus, the amplitude of the given transverse wave in the rope is $\boxed{0.750\,{\text{cm}}}$.

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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.