1. The speed of a transverse wave on a stretched string is given by

$v = \sqrt{\frac{T}{\mu}}$

where:

• $T$ = tension in the string
• $\mu$ = linear mass density

2. Here, the tension is increased by $4\%$.

So the new tension is

$T' = 1.04T$

3. The new wave speed becomes

$v' = \sqrt{\frac{T'}{\mu}} = \sqrt{\frac{1.04T}{\mu}} = \sqrt{1.04}\,\sqrt{\frac{T}{\mu}}$

Thus,

$v' = \sqrt{1.04}\,v$

4. Now,

$\sqrt{1.04} \approx 1.0198$

So the fractional increase in speed is

$\frac{v' - v}{v} = 1.0198 - 1 = 0.0198$

5. Converting to percentage:

$0.0198 \times 100 = 1.98\%$

which is approximately

$2\%$

6. Therefore, the percentage increase in speed is

$\boxed{2}$