SATHEE: Wave Motion 2 Question 8

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 $100 H z$ . The separation between successive nodes on the string is close to

(2019 Main, 10 Jan I)

(a) $16.6 c m$

(b) $33.3 c m$

(d) $20.0 c m$

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Answer:

Correct Answer: 8. (d)

Solution:

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

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

Substituting the given values, we get

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

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

where, $f =$ frequency of wave.

$\Rightarrow λ = \frac{40}{100} m = 40 c m$

$∴$ Separation between two successive nodes is,

$d = \frac{λ}{2} = \frac{40}{2} = 20.0 c m$

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Wave Motion 2 Question 8

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

Answer:

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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 $100 H z$ . The separation between successive nodes on the string is close to