Physics 2020 JAMB Past Questions - Question 19: The lowest note emitted by a streched string has a frequency of 40 Hz. How many overtones are there bertween 40Hz and 180 Hz?

Correct Answer

D

Explanation

F₁ = 3F₀, F₂ = 5F₀ ; F₁ = 120HZ F₂ = 200 HZ i.e 1overtone
The lowest note emitted by a stretched string, also known as the fundamental frequency, sets the stage for the overtones. The overtones are integer multiples of the fundamental frequency. To find the number of overtones between 40 Hz and 180 Hz, we can calculate the overtones of the fundamental frequency and see how many fall within the given range.

The overtones can be calculated using the formula:
\[ f_n = nf_1 \]
where \( f_n \) is the frequency of the nth overtone and \( f_1 \) is the fundamental frequency.

So, the overtones are:
\[ f_2 = 2 \times 40 \, \text{Hz} = 80 \, \text{Hz} \]
\[ f_3 = 3 \times 40 \, \text{Hz} = 120 \, \text{Hz} \]
\[ f_4 = 4 \times 40 \, \text{Hz} = 160 \, \text{Hz} \]

Therefore, there are 3 overtones between 40 Hz and 180 Hz.