A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to:

We first note that a transverse wave on a stretched string travels with a speed given by the formula

$$v=\sqrt{\dfrac{T}{\mu}},$$

where $$T$$ is the tension in the string and $$\mu$$ is the linear mass density.

The string has length $$L = 1\ \text{m}$$ and mass $$m = 5\ \text{g} = 0.005\ \text{kg}$$, so

$$\mu=\dfrac{m}{L}=\dfrac{0.005\ \text{kg}}{1\ \text{m}} = 0.005\ \text{kg\,m}^{-1}.$$

Substituting $$T = 8\ \text{N}$$ and $$\mu = 0.005\ \text{kg\,m}^{-1}$$ into the speed formula, we have

$$v=\sqrt{\dfrac{8}{0.005}}=\sqrt{1600}=40\ \text{m\,s}^{-1}.$$

A string fixed at both ends supports standing waves whose natural frequencies are

$$f_n=\dfrac{n v}{2L},\qquad n=1,2,3,\ldots$$

An external vibrator of frequency $$f = 100\ \text{Hz}$$ is applied. Setting this equal to one of the natural frequencies, we get

$$100 = \dfrac{n\,(40)}{2\,(1)} = 20n \quad\Longrightarrow\quad n = 5.$$

For the $$n^{\text{th}}$$ harmonic, the wavelength is

$$\lambda_n=\dfrac{2L}{n} = \dfrac{2\,(1\ \text{m})}{5}=0.4\ \text{m}=40\ \text{cm}.$$

The distance between successive nodes on a standing wave is one-half of the wavelength, i.e.

$$\text{node separation}=\dfrac{\lambda_n}{2}=\dfrac{0.4\ \text{m}}{2}=0.2\ \text{m}=20\ \text{cm}.$$

Hence, the correct answer is Option A.