Two mechanical waves on strings of equal length (L) tension (T) having linear mass densities μ₁/μ₂ = 1/2. Find the ratio of time taken for a wave pulse to travel from one end to the other in both strings. (Ignore gravity)

To find the ratio of the time taken for a wave pulse to travel from one end to the other in both strings, we start with the formula for the speed of a transverse wave on a stretched string:
$v=\sqrt{\frac{T}{\mu }}$
where:
- $T$ is the tension in the string,
- $\mu$ is the linear mass density of the string.

The time $t$ taken by the wave pulse to travel the length $L$ of the string is given by:
$t=\frac{L}{v}$

Substituting the expression for speed $v$ into the time equation:
$t=\frac{L}{\sqrt{\frac{T}{\mu }}}$
Simplifying this expression gives:
$t=L\sqrt{\frac{\mu }{T}}$

Since both strings have the same length ($L$) and the same tension ($T$), the time taken $t$ is directly proportional to the square root of the linear mass density $\mu$:
$t\propto \sqrt{\mu }$

Therefore, the ratio of the time taken in both strings ($t_1/t_2$) is:
$\frac{t_1}{t_2}=\sqrt{\frac{{\mu }_1}{{\mu }_2}}$

Given that the ratio of the linear mass densities is $\frac{{\mu }_1}{{\mu }_2}=\frac{1}{2}$, we substitute this value into our ratio equation:
$\frac{t_1}{t_2}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$

Thus, the ratio of the time taken for the wave pulse to travel in both strings is 1/√2.