Separation Between Successive Nodes on Vibrating Stretched String

To find the separation between successive nodes on the vibrating string, we follow these steps:

1. Calculate the Linear Mass Density ($\mu$):
   The mass of the string is $m = 1.0\text{ g} = 1.0 \times 10^{-3}\text{ kg}$, and its length is $L = 20\text{ cm} = 0.20\text{ m}$.
   $\mu = \frac{m}{L} = \frac{1.0 \times 10^{-3}\text{ kg}}{0.20\text{ m}} = 5.0 \times 10^{-3}\text{ kg/m}$

2. Calculate the Speed of Transverse Waves ($v$):
   The tension in the string is $T = 0.5\text{ N}$. The wave speed $v$ is given by:
   $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{0.5\text{ N}}{5.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{100} = 10\text{ m/s}$

3. Calculate the Wavelength ($\lambda$):
   The string is driven by an external vibrator of frequency $f = 100\text{ Hz}$. The wavelength $\lambda$ of the wave formed on the string is:
   $\lambda = \frac{v}{f} = \frac{10\text{ m/s}}{100\text{ Hz}} = 0.10\text{ m} = 10\text{ cm}$

4. Calculate the Separation Between Successive Nodes ($d$):
   In a standing wave, the distance between any two consecutive nodes is equal to half of the wavelength:
   $d = \frac{\lambda}{2} = \frac{10\text{ cm}}{2} = 5\text{ cm}$

(Note: The fundamental frequency of the string is $f_1 = \frac{v}{2L} = \frac{10}{2 \times 0.20} = 25\text{ Hz}$. Since the vibrator frequency $100\text{ Hz}$ is an integer multiple of $f_1$ ($n = 4$), the string vibrates in resonance in its 4th harmonic with 4 loops, each having a length of $5\text{ cm}$.)

Thus, the separation between successive nodes is 5 cm.