Question

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

Short answer

The tension in the string is approximately 6.25 N. $T\approx 6.25N$

Step-by-step solution

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): $0.04kg/m$

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: $v=\frac{2\pi (1/0.04)}{1/0.50}=\frac{2\pi (50)}{2}=50\pi$

Calculate tension using linear mass density and wave speed

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

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

Key concepts

Wave Equation

The wave equation is a formula that describes the motion of a wave through a medium, like a string. A typical wave equation for waves on a string is given by $y(x,t)=A\sin (kx-\omega t)$. Here is what each symbol stands for:

• $y$ represents the displacement at a position $x$ and time $t$.
• $A$ is the amplitude, which is the maximum displacement of the wave from its rest position.
• $k$ is the wave number, calculated as $k=\frac{2\pi }{\lambda }$, where $\lambda$ is the wavelength.
• $\omega$ is the angular frequency, which we'll discuss further below.

In the given problem, the wave equation is: $y=0.02\sin [2\pi (1/0.04)-(x/0.50)]$. This indicates both the wave's amplitude and its behavior through the string. Exploring this equation helps us understand how quickly and in what manner the wave propagates along the string.

Linear Mass Density

Linear mass density, often symbolized as $\mu$, refers to the mass of the string per unit of length. It is a vital parameter for understanding how a wave behaves as it travels through a string. For this exercise, the linear mass density is given as $0.04kg/m$.
Linear mass density influences the wave's speed and the tension in the string significantly. To determine how fast a wave travels, or the force exerted on a string, knowing the linear mass density is crucial.When we combine this with the wave speed, we can find the tension using the relationship $v=\sqrt{\frac{T}{\mu }}$, where $T$ is the tension and $v$ is the wave speed.

Wave Speed

Wave speed is how fast the wave travels down the string. It's crucial for determining characteristics like tension. In the general wave equation $y(x,t)=A\sin (kx-\omega t)$, the wave speed $v$ is related to angular frequency $\omega$ and wave number $k$ by the formula:$$v=\frac{\omega }{k}\text{.}$$In the example given, we have:

• $\omega =2\pi (1/0.04)$
• $k=1/0.50$

Using the given values, plug them into the wave speed formula to obtain:$$v=\frac{2\pi (50)}{2}=50\pi$$This calculation shows that the wave travels at 50\pi meters per second along the string.

Angular Frequency

Angular frequency, denoted by $\omega$, measures how quickly the wave oscillates in radians per second. It's a key factor in understanding the dynamics of wave motion. In a wave equation, $\omega$ is part of what determines both wave speed and wave energy.In mathematical terms, angular frequency is calculated as:$$\omega =2\pi f$$where $f$ is the frequency in cycles per second. In the context of our exercise, we used $\omega =2\pi (1/0.04)$ to find its precise value for the wave in the string.Understanding angular frequency allows us to link it with wave number $k$ for practical applications, such as calculating wave speed or assessing the wave's effect on string tension.