The time period of the transverse wave in a rope will be [tex]\boxed{{\text{0}}{\text{.025 s}}}[/tex].
Further Explanation:
The waves are the disturbance created in the medium particles. A transverse wave in a rope is the disturbance produced in the particles of the rope in the vertical direction.
A transverse wave is a wave in which the disturbance in the medium is perpendicular to the propagation of the wave. The time period of the wave is the time taken in completing one oscillation about the mean position.
The expression of the transverse wave in the rope is given.
[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 for a wave present in a rope is given as.
[tex]\boxed{y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)}[/tex]
Here, [tex]A[/tex] is the amplitude of the wave, [tex]y[/tex] is the position of the wave, [tex]k[/tex] is the wave number and [tex]\omega[/tex] is the angular frequency of the wave.
From the above expression, the angular frequency of the wave is obtained as.
[tex]\omega = 250\pi[/tex]
We know that the time period is obtained from the angular frequency as.
[tex]T = \dfrac{{2\pi }}{\omega }[/tex]
Substitute the value of [tex]\omega[/tex] in above expression.
[tex]\begin{aligned}T &= \frac{{2\pi }}{{250\pi }} \\&= \frac{1}{{125}}\,{\text{s}} \\&\approx {\text{0}}{\text{.008}}\,{\text{s}} \\&\approx {\text{8}}\,{\text{ms}} \\\end{aligned}[/tex]
Thus, the time period of the transverse wave in the rope is [tex]\boxed{8\,{\text{ms}}}[/tex].
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Answer Details:
Grade: High School
Chapter: Waves
Subject: Physics
Keywords: Transverse, wave, amplitude, angular frequency, perpendicular, stationary wave, rope, velocity, position, propagation of wave, wave number.
