Question

A string of length 1 m 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
1. 16.6 cm
2. 33.3 cm
3. 10.0 cm
4. 20.0 cm

Answer 1

Correct Answer - Option 4 : 20.0 cm

Concept:

Velocity 'v' of the wave on the string is given by,

\(v = \sqrt {\frac{T}{\mu }}\)

Where, T = tension and μ = mass per unit length

Wavelength of the wave on the string,

\(\lambda = \frac{v}{f}\)

Calculation:

Given,

Length of the string, l = 1 m

Mass of the string, m = 5 g

String Tension = 8.0 N

Frequency of the vibration = 100 Hz

\(\Rightarrow \mu = \frac{m}{l} = \frac{{5\;g}}{{1\;m}} = 5\;g\)

\(\Rightarrow \mu = 5 \times {10^{ - 3}}Kg = \frac{5}{{1000}}kg\)

Substituting the given values, we get,

\(\Rightarrow v = \sqrt {\frac{8}{{\left( {\frac{5}{{1000}}} \right)}}} = \sqrt {8 \times \frac{{1000}}{5}} \)

\(\Rightarrow v = \sqrt {8 \times 200} = \sqrt {1600} \)

∴ v = 40 ms^-1

Wavelength of the wave on the string,

\({\rm{\lambda }} = \frac{{\rm{v}}}{{\rm{f}}}\)

Where, f = frequency of wave

\(\Rightarrow \lambda = \frac{{40}}{{100}}m = 0.4\;m\)

⇒ λ = 0.4 × 10² cm

∴ λ = 40 cm

∴ Separation between two successive nodes is,

\(\Rightarrow d = \frac{\lambda }{2} = \frac{{40}}{2}\)

∴ d = 20.0 cm

Therefore, the separation between successive nodes on the string is close to 20.0 cm.