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Chapter 9: Problem 246

A rope, under tension of \(200 \mathrm{~N}\) and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by \(y=(0.10 \mathrm{~m}) \sin \left(\frac{\pi x}{2}\right) \sin (12 \pi t)\), where \(x=0\) at one end of the rope, \(x\) is in metres, and \(t\) is in seconds. Find (A) the length of the rope (B) the speed of waves on the rope (C) the mass of the rope (D) if the rope oscillates in a third harmonic standing wave pattern, what will be the period of oscillation?

Short Answer

Expert verified
(A) The length of the rope is \(L = 2 \mathrm{~m}\). (B) The speed of waves on the rope is \(v = 12 \mathrm{~m/s}\). (C) The mass of the rope is \(M = \frac{100}{36} \mathrm{~kg}\). (D) The period of oscillation for the third harmonic is \(T = \frac{1}{2} \mathrm{~s}\).

Step by step solution

01

Find the length of the rope

To find the length of the rope, we need to understand the relationship between the displacement equation and the second harmonic of a standing wave. For a second harmonic, there are two loops in the wave pattern. Since the displacement equation is given by: \(y=(0.10 \mathrm{~m}) \sin \left(\frac{\pi x}{2}\right) \sin (12 \pi t)\) We can use the sine term with the variable x, which is \(\sin \left(\frac{\pi x}{2}\right)\) to find the length of the rope (L). The equation represents half a complete cycle, so when x = L, the argument of the sine function must be equal to \(\pi\): \(\frac{\pi x}{2} = \pi\) Now solve for x (L): \(L = \frac{2\pi}{\pi}\)
02

Calculate the length of the rope

Now that we have the relationship between x and the length of the rope, we can calculate the length (L) as follows: \(L = \frac{2\pi}{\pi} = 2 \mathrm{~m}\)
03

Determine the speed of waves on the rope

To find the speed of the waves (v) on the rope, we can use the fact that the time-dependent part of the displacement equation is \(\sin (12 \pi t)\), which represents the angular frequency \(\omega = 12 \pi\). The speed of the wave can be found using the relationship between speed (v), angular frequency (\(\omega\)), and the wavelength (\(\lambda\)): \(v = \frac{\omega \lambda}{2 \pi}\) Since we know the wavelength represents one complete cycle for the second harmonic and the length of the rope is 2 meters (2 loops in the wave pattern), we can find the speed of the waves using the given angular frequency: \(v = \frac{12\pi (2)}{2\pi}\)
04

Calculate the speed of waves on the rope

Now that we have the relationship to find the speed, we can calculate the speed (v): \(v = \frac{12\pi (2)}{2\pi} = 12 \mathrm{~m/s}\)
05

Calculate the mass of the rope

To find the mass of the rope, we can use the wave speed equation and the tension (T) given in the problem: \(v = \sqrt{\frac{T}{\mu}}\) Where \(\mu\) represents the mass per unit length of the rope. Rearranging the equation to solve for \(\mu\): \(\mu = \frac{T}{v^2}\) Now, we can calculate the mass per unit length using the given tension (T = 200 N) and the calculated wave speed (v = 12 m/s): \(\mu = \frac{200}{(12)^2}\) Finally, to find the total mass (M) of the rope, we multiply the mass per unit length by the length of the rope (L = 2 m): \(M = \mu \times L\)
06

Find the mass of the rope

Calculating the total mass (M) of the rope: \(M = \frac{200}{(12)^2} \times 2 = \frac{100}{36} \mathrm{~kg}\)
07

Find the period of oscillation for the third harmonic

When the rope oscillates in a third harmonic standing wave pattern, the relationship between wavelength (\(\lambda\)) and length (L) becomes: \(L = \frac{3}{2} \lambda\) Since we found that the speed of the waves on the rope remains constant, we can use this fact to solve for the period (T) of the third harmonic oscillation. The speed of the waves is given by: \(v = \frac{\lambda}{T}\) We can rewrite the equation to solve for the period (T) using the length of the rope (L = 2 m) and the relationship with the wavelength from the third harmonic pattern: \(T = \frac{2(\frac{3}{2})\lambda}{12}\)
08

Calculate the period of oscillation

Now that we have the relationship to find the period, we can calculate the period (T) of the third harmonic oscillation: \(T = \frac{2(\frac{3}{2})(2)}{12} = \frac{1}{2} \mathrm{~s}\) #Results#: Using the given exercise and our analysis, we have found the following values: (A) The length of the rope: \( L = 2 \mathrm{~m}\) (B) The speed of waves on the rope: \( v = 12 \mathrm{~m/s}\) (C) The mass of the rope: \( M = \frac{100}{36} \mathrm{~kg}\) (D) The period of oscillation for the third harmonic: \( T = \frac{1}{2} \mathrm{~s}\)

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Most popular questions from this chapter

Two monochromatic coherent point sources \(S_{I}\) and \(S_{2}\) are separated by a distance \(L .\) Each source emits light of wavelength \(\lambda\), where \(L \gg \lambda\). The line \(S_{1} S_{2}\) when extended meets a screen perpendicular to it at a point \(A\). (A) The interference fringes on the screen are circular in shape. (B) The interference fringes on the screen are straight lines perpendicular to the line \(S_{1} S_{2} A\). (C) The point \(A\) is an intensity maxima if \(L=n \lambda\). (D) The point \(A\) is always an intensity maxima for any separation \(L\).

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