A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is:

We have a stretched string of length $$L = 2.0\ \text{m}$$, and both its ends are fixed. Such a string supports standing waves whose frequencies are integral multiples of the fundamental frequency. If the string is vibrating in its third harmonic (that is, the third normal mode), the frequency supplied by the vibrator is the third-harmonic frequency $$f_{3}$$.

By definition of harmonics for a string fixed at both ends, the relation between the harmonic number $$n$$ and the corresponding frequency $$f_{n}$$ is

$$f_{n} = n\,f_{1},$$

where $$f_{1}$$ is the fundamental frequency (also called the first harmonic). For the third harmonic we set $$n = 3$$, so

$$f_{3} = 3\,f_{1}.$$

The vibrator is known to supply $$f_{3} = 240\ \text{Hz}$$. Substituting this value, we get

$$240\ \text{Hz} = 3\,f_{1}.$$

Now, solving for $$f_{1}$$,

$$f_{1} = \frac{240\ \text{Hz}}{3} = 80\ \text{Hz}.$$

So, the fundamental frequency of the string is $$80\ \text{Hz}.$$

Next, we need the speed $$v$$ of the wave along the string. For any wave, the basic relation connecting speed $$v$$, frequency $$f$$ and wavelength $$\lambda$$ is

$$v = f\,\lambda.$$

For a string fixed at both ends, the wavelength of the $$n^{\text{th}}$$ harmonic is determined by the formula

$$\lambda_{n} = \frac{2L}{n}.$$

Putting $$n = 3$$ for the third harmonic and $$L = 2.0\ \text{m}$$, we find

$$\lambda_{3} = \frac{2 \times 2.0\ \text{m}}{3} = \frac{4.0\ \text{m}}{3} = 1.\overline{3}\ \text{m}.$$

Now insert $$f_{3} = 240\ \text{Hz}$$ and $$\lambda_{3} = \dfrac{4}{3}\ \text{m}$$ into the wave-speed relation:

$$v = f_{3}\,\lambda_{3} = 240\ \text{Hz} \times \frac{4}{3}\ \text{m}.$$

Carrying out the multiplication,

$$v = 240 \times 1.\overline{3}\ \text{m s}^{-1} = 320\ \text{m s}^{-1}.$$

Thus, the speed of the wave on the string is $$320\ \text{m s}^{-1},$$ and the fundamental frequency is $$80\ \text{Hz}.$$

Hence, the correct answer is Option B.