Minimum and maximum tension in transverse wave

The tension of the string is a constant, if there is no vibration on the string. A wave is produced on the string when you give an unbalanced force on the string which varies the original tension of the string. The velocity of the wave now depends on the value of the tension. The given equation is valid only for small amplitude vibrations.

The tension is minimum and a constant, when no wave propagates through it. i.e., the velocity of the wave is zero. Now when a wave propagates through the medium, the tension increases and depends on the velocity (actually the frequency) of the wave. Find the tension at zero velocity and the velocity provided in the eqn. of the wave in your question. Find their ratio which will give you the required answer.

When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it.

If we take a segment small enough that we can consider it as a straight line, then the oscillation has stretched it from its rest length of $ds$ to the length $ds$. So if the rest tension is $T$ then the tension in the stretched segment is:

$$ T' = T \frac{ds}{dx} $$

So calculating the tension reduces to finding the ratio $ds/dx$. This can be related to the angle $\theta$ by:

$$ \frac{dx}{ds} = \cos\theta $$

and the angle $\theta$ is related to the gradient by:

$$ \frac{dy}{dx} = \tan\theta $$