JEE MAIN - Physics (2023 - 8th April Evening Shift - No. 19)
Explanation
The fundamental frequency (also known as the first harmonic) of a vibrating string is given by the formula:
$f = \frac{v}{2L}$
where:
- (f) is the frequency,
- (v) is the speed of the wave in the string, and
- (L) is the length of the string.
In this case, the speed of the wave in the string stays the same because it depends on the properties of the string and the tension in it, which we can assume to be constant.
We can write the equation for the fundamental frequency of the original string and the shorter string:
$f_1 = \frac{v}{2L_1}$
$f_2 = \frac{v}{2L_2}$
where:
- $(f_1 = 120 \, \text{Hz})$ and $ (L_1 = 90 \, \text{cm})$ for the original string, and
- $(f_2 = 180 \, \text{Hz})$ and $(L_2)$ is what we're trying to find for the shorter string.
We can set up a ratio of these two equations:
$\frac{f_1}{f_2} = \frac{L_2}{L_1}$
Substituting in the given values, we get:
$\frac{120 \, \text{Hz}}{180 \, \text{Hz}} = \frac{L_2}{90 \, \text{cm}}$
Solving for ($L_2$) gives:
$L_2 = 90 \, \text{cm} \times \frac{120 \, \text{Hz}}{180 \, \text{Hz}} = 60 \, \text{cm}$
So, the length of the string producing a fundamental frequency of 180 Hz will be 60 cm.
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