Transverse wave equation period

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To find the period of the transverse wave described by the equation y(x, t) = (0.750 cm) cos(π[(0.400 cm⁻¹)x + (250 s⁻¹)t]), it is necessary to rewrite it in standard form. The standard form is y(x, t) = Acos[2π(x/λ + t/T)], where A is the amplitude, λ is the wavelength, and T is the period. By comparing the coefficients in the given equation, the angular frequency can be identified as ω = 250 s⁻¹. The period T can then be calculated using the relationship T = 2π/ω, resulting in T = 0.025 s. Understanding this process is crucial for solving similar problems in physics.
james brug
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A transverse wave on a rope is given by y(x, t)=<br /> (0.750\; {\rm cm})\, \cos ( \, \pi [(0.400\;{\rm cm}^{ - 1})x+(250\; {\rm s}^{ - 1})t])


Find the period.

This should be simple, but I keep getting the wrong answer in Mastering Physics. I can't find any explanation in my book, and it's really irritating me.
 
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Rewrite the equation in a standard form.
y(x,t) = Acos[2*pi(x/lambda + t/T)]
Compare this with the given equation and find the period.
 

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