Wave Motion 2 Question 8

8. 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

(2019 Main, 10 Jan I)

(a) $16.6 cm$

(b) $33.3 cm$

(c) $10.0 cm$

(d) $20.0 cm$

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Answer:

Correct Answer: 8. (d)

Solution:

  1. Velocity ’ $v$ ’ of the wave on the string $=\sqrt{\frac{T}{\mu}}$

where, $T=$ tension and $\mu=$ mass per unit length.

Substituting the given values, we get

$$ v=\sqrt{\frac{8}{5} \times 1000}=40 ms^{-1} $$

Wavelength of the wave on the string, $\lambda=\frac{v}{f}$

where, $\quad f=$ frequency of wave.

$$ \Rightarrow \quad \lambda=\frac{40}{100} m=40 cm $$

$\therefore$ Separation between two successive nodes is,

$$ d=\frac{\lambda}{2}=\frac{40}{2}=20.0 cm $$



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