The classic string-and-pulley setup ties the wave speed of Wave Speed in a Medium to the loops of Standing Waves on Strings. The hanging mass sets the tension. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
$$T = Mg, \qquad v = \sqrt{\frac{T}{\mu}}, \qquad L = n\frac{\lambda}{2}, \qquad f = \frac{v}{\lambda} = \frac{nv}{2L}.$$ At a fixed frequency, $n \propto 1/v \propto 1/\sqrt{M}$.
Worked example: Three loops
A string with $\mu = 2.0\ \text{g/m}$ runs $1.2\ \text{m}$ to a pulley holding $1.0\ \text{kg}$ and shows 3 loops. Find (1) the wave speed and (2) the frequency.
- (1) $v = \sqrt{(1.0)(9.8)/0.002} = 70.0\ \text{m/s}$.
- (2) $\lambda = 2(1.2)/3 = 0.8\ \text{m}$, so $f = 70.0/0.8 = 87.5\ \text{Hz}$.
Answer: $70.0\ \text{m/s}$, $87.5\ \text{Hz}$
Common mistake
Treating speed as proportional to tension. It goes as the square root: 4 times the hanging mass only doubles the speed. Check units: $\sqrt{\text{N}/(\text{kg/m})} = \text{m/s}$.