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Chapter 10: Problem 1431

If the equation of a wave in a string having linear mass density $0.04 \mathrm{~kg} \mathrm{~m}^{-1}\( is given by \)\mathrm{y}=0.02\( \)\sin [2 \pi\\{1 /(0.04)\\}-\\{\mathrm{x} /(0.50)\\}]$, then the tension in the string is \(\ldots \ldots \ldots \ldots\) N. (All values are in \(\mathrm{mks}\) ) (A) \(6.25\) (B) \(4.0\) (C) \(12.5\) (D) \(0.5\)

Short Answer

Expert verified
The tension in the string is approximately 6.25 N. \(T \approx 6.25\: N\)

Step by step solution

01

Identify the given values and the wave equation

We are given the wave equation and linear mass density: Wave equation: \(y = 0.02 \sin[2\pi(1/0.04) - (x/0.50)]\) Linear mass density (µ): \(0.04\: kg/m\)
02

Determine the wave speed from the wave equation

To determine the wave speed, we need to look at the factor that multiplies x and extract its value. The general equation for a wave on a string is: \(y(x, t) = A\sin(kx - \omega t)\) Where A is the amplitude, k is the wave number, x is the position, t is the time, and ω is the angular frequency. From the given wave equation, we know the angular frequency is \(2\pi(1/0.04)\), and the wave number is \(1/0.50\). We can relate these two quantities to the wave speed (v) using the formula: \(v = \frac{\omega}{k}\) Substitute the values of ω and k to find the wave speed: \(v = \frac{2\pi(1/0.04)}{1/0.50} = \frac{2\pi(50)}{2} = 50\pi\)
03

Calculate tension using linear mass density and wave speed

Use the formula relating tension (T), wave speed (v), and linear mass density (µ): \(v = \sqrt{\frac{T}{µ}}\) Solve for tension (T): \(T = µv^2\) Substitute the values of µ and v into the formula: \(T = 0.04 \times (50\pi)^2\) Calculate the tension: \(T \approx 6.25\: N\)
04

Determine the correct answer

The tension in the string is approximately 6.25 N, which corresponds to option (A). So the correct answer is: (A) 6.25 N

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