Question

The linear density of a vibrating string is 1.3 × 10^-4 kg m1. A transverse wave is propagating on the string and is described by the equation y (x, t) = 0.021 sin (30t - x) where x and y are in metres and t is in seconds. Calculate the tension in the string.

Answer 1

Answer on Question #65640, Physics / Mechanics | Relativity

The linear density of a vibrating string is $1.3 \times 10^{\wedge} - 4\ \mathrm{kg\ m^{-3}}$. A transverse wave is propagating on the string and is described by the equation $y(x,t) = 0.021\sin(30t - x)$ where $x$ and $y$ are in metres and $t$ is in seconds. Calculate the tension in the string.

Find: $T - ?$

Given:

$$\begin{array}{l} \rho = 1.3 \times 10^{-4}\ \mathrm{kg} \times \mathrm{m}^{-3} \\ y(x,t) = 0.021\sin(30t - x) \\ \end{array}$$

Solution:

Wave velocity:

$$v = \sqrt{\frac{T}{\rho}}(1), \text{ where } T \text{ is tension, } \rho \text{ is density} \\ \text{Of } (1) \Rightarrow v^2 = \frac{T}{\rho}(2) \\ \text{Of } (2) \Rightarrow T = v^2 \times \rho(3) \\$$

Wave velocity:

$$v = f \times \lambda(4)$$

Linear frequency:

$$f = \frac{\omega}{2\pi}(5), \text{ where } \omega \text{ is cyclic frequency}$$

Wavelength:

$$\lambda = \frac{2\pi}{k}(6), \text{ where } k \text{ is wave number}$$

(5) and (6) in (4):

$$v = \frac{\omega}{2\pi} \times \frac{2\pi}{k} = \frac{\omega}{k}(7)$$

Equation of plane wave:

$$y(x,t) = A \sin(\omega t - x)(8)$$

From the condition of the task:

$$y(x,t) = 0.021\sin(30t - x)(9)$$

Of (8) and (9) $\Rightarrow \omega = 30\ \mathrm{s}^{-1},\ k = 1\ \mathrm{m}^{-1}(10)$

(10) in (7): $v = 30\ \mathrm{m} \times \mathrm{s}^{-1}(11)$

(11) in (3) $T = 0.117\ \mathrm{N}$

Answer:

0.117 N

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