JAMB Physics 2023 Paper

Concept:For a stretched string, the fundamental frequency depends on its length, tension, and linear density.Each overtone is a whole-number multiple of the fundamental frequency.Explanation:Convert all values to SI units first.Length: $L = 50\,\text{cm} = 0.50\,\text{m}$.Mass: $m = 10\,\text{g} = 0.010\,\text{kg}$.Linear density: $\mu = \frac{m}{L} = \frac{0.010}{0.50} = 0.020\,\text{kg/m}$.The fundamental frequency of a string is given by:$$f_0 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$Substitute $L = 0.50\,\text{m}$, $T = 800\,\text{N}$, and $\mu = 0.020\,\text{kg/m}$:$$f_0 = \frac{1}{2(0.50)}\sqrt{\frac{800}{0.020}}$$$$f_0 = \frac{1}{1.00}\sqrt{40000} = 200\,\text{Hz}$$The third overtone is the fourth harmonic.So its frequency is:$$f = 4f_0 = 4 \times 200 = 800\,\text{Hz}$$Answer:$800\,\text{Hz}$ $\Rightarrow$ Option A.