Animated Solution for Physics - Waves: The equation of a wave on a string of linear mass density $0.04{\text{ kg m}}^{-1}$ is given by $y=0.02\text{ (m)}\sin \left[2\pi \left(\frac{t}{0.04\text{ (s)}}-\frac{x}{0.50\text{ (m)}}\right)\right]$. The tension in the string is

The beauty of wave mechanics lies in how a simple mathematical equation can perfectly describe a dynamic physical reality. In this problem, we are given the equation of a transverse wave traveling on a string and asked to find the physical tension that sustains it. This requires us to bridge the gap between the kinematics of the wave (how it moves) and the dynamics of the string (the forces involved).

We are given the wave equation:

At first glance, this might look a bit intimidating because of the fractions and the $2\pi$ sitting outside the bracket. However, the key to unlocking this problem is to compare it with the standard equation of a traveling wave:

Here, $A$ represents the amplitude, $\omega$ is the angular frequency, and $k$ is the wave number. To make a direct comparison, we must distribute the $2\pi$ into the bracket.

By distributing the $2\pi$, our given equation becomes:

Now, the comparison is straightforward. We can immediately extract the vital kinematic parameters of the wave:

Angular Frequency: $\omega =\frac{2\pi }{0.04}\text{ rad/s}$

Wave Number: $k=\frac{2\pi }{0.50}\text{ rad/m}$

The speed at which a wave propagates through a medium, denoted by $v$, is fundamentally linked to its angular frequency and wave number. The relationship is given by the simple ratio:

Let's substitute the values we just extracted:

Notice how beautifully the $2\pi$ terms cancel out in the numerator and denominator. This leaves us with:

So, the wave is zipping along the string at a brisk $12.5\text{ m/s}$. But we aren't done yet; we need to find the tension.

This is where we transition from kinematics to dynamics. The speed of a transverse wave on a stretched string is determined entirely by the physical properties of the string: the tension $T$ pulling it taut, and its linear mass density $\mu$ (mass per unit length). The classic formula derived from Newton's laws is:

This equation tells a profound physical story. The tension $T$ provides the restoring force; a tighter string snaps back faster, leading to a higher wave speed. Conversely, the linear mass density $\mu$ represents inertia; a heavier, thicker string is more sluggish, leading to a lower wave speed.

We need to solve for the tension $T$. Let's square both sides of our dynamic equation to isolate $T$:

We are given the linear mass density $\mu =0.04\text{ kg/m}$, and we just calculated the wave speed $v=12.5\text{ m/s}$. Substituting these values in:

And there we have it! The tension required to maintain this specific wave on the string is exactly $6.25\text{ N}$. This problem elegantly demonstrates how mathematical wave parameters are intimately tied to the physical forces acting on the medium.