Let's analyze the problem step by step:

For a string fixed at both ends, the resonant (harmonic) frequencies are given by

$$ f_n = \frac{n v}{2L}, $$

where

$$v$$ is the speed of the transverse waves,

$$L$$ is the length of the string (here, 0.8 m),

$$n$$ is the mode number (an integer: 1, 2, 3, …).

Notice that the given resonant frequencies are 90 Hz, 150 Hz, and 210 Hz. They are equally spaced:

150 Hz − 90 Hz = 60 Hz,

210 Hz − 150 Hz = 60 Hz.

This constant spacing indicates that the difference between two consecutive modes is

$$ \Delta f = \frac{v}{2L} = 60 \text{ Hz}. $$

However, if the spacing between the harmonics is 60 Hz, the fundamental frequency for a string fixed at both ends would be

$$ f_1 = \frac{v}{2L} = 60 \text{ Hz}. $$

But then the harmonics would be 60 Hz, 120 Hz, 180 Hz, … which do not match our given frequencies.

To resolve this, we consider that the observed frequencies might not start from the first harmonic. Suppose the observed frequencies correspond to the 3rd, 5th, and 7th modes. Let’s test this:

For the 3rd harmonic, we have

$$ f_3 = \frac{3v}{2L}. $$

Given that $$f_3 = 90 \text{ Hz}$$, we solve for $$v$$:

$$ 90 = \frac{3v}{2(0.8)} \quad \Longrightarrow \quad v = \frac{90 \times 2 \times 0.8}{3}. $$

Calculate the wave speed:

First calculate the numerator: $$90 \times 2 \times 0.8 = 90 \times 1.6 = 144.$$

Now divide by 3:

$$ v = \frac{144}{3} = 48 \text{ m/s}. $$

Checking with the other frequencies:

For $$n = 5$$:

$$ f_5 = \frac{5v}{2L} = \frac{5 \times 48}{2 \times 0.8} = \frac{240}{1.6} = 150 \text{ Hz}, $$

For $$n = 7$$:

$$ f_7 = \frac{7v}{2L} = \frac{7 \times 48}{1.6} = \frac{336}{1.6} = 210 \text{ Hz}. $$

Both calculations match the given resonant frequencies.

Therefore, the speed of the transverse wave in the string is $$48 \text{ m/s}$$, which corresponds to Option C.

Waves

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