First slide

Stationary waves

Question

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

Moderate

A

16.6 cm
B

20.0 cm
C

10.0 cm
D

33.3 cm

Solution

$V e l o c i t y o f w a v e o n s t r i n g V = \sqrt{\frac{T}{μ}} = \sqrt{\frac{8}{5 \times 10^{- 3}}} = 40 m / \sec W a v e l e n g t h o f w a v e : λ = \frac{v}{f} \Rightarrow λ = \frac{40}{100} m S e p a r a t i o n b e t w e e n s u c c e s s i v e n o d e s = λ / 2 = \frac{20}{100} = 0.2 m = 20 c m$

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