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Chapter 11: Problem 24

Write the equation for a transverse sinusoidal wave with a maximum amplitude of \(2.50 \mathrm{cm}\) and an angular frequency of 2.90 rad/s that is moving along the positive \(x\) -direction with a wave speed that is 5.00 times as fast as the maximum speed of a point on the string. Assume that at time \(t=0,\) the point \(x=0\) is at \(y=0\) and then moves in the \(-y\) -direction in the next instant of time.

Short Answer

Expert verified
\(\lambda = 10\pi A\) So the wavelength of the transverse sinusoidal wave is equal to 10 times the amplitude multiplied by pi.

Step by step solution

01

Find the wavelength#

Since the wave speed is 5.00 times the maximum speed of a point on the string, we can use this information to find the wavelength. The maximum speed of a point on the string is given by the product of amplitude and angular frequency: \(v_{max} = A \cdot \omega\) So the wave speed, v, is related to the maximum speed as: \(v = 5.00 \cdot v_{max} = 5.00 \cdot (A \cdot \omega)\) The wave speed is also related to the wavelength and angular frequency through the equation: \(v=\frac{\omega \cdot \lambda}{2\pi}\) We can now solve for the wavelength: \(\lambda = \frac{2\pi\cdot v}{\omega} = \frac{2\pi\cdot 5.00 \cdot (A \cdot \omega)}{\omega}\)

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Most popular questions from this chapter

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