Wave Motion 2 Question 40
40. A $20 cm$ long string, having a mass of $1.0 g$, is fixed at both the ends. The tension in the string is $0.5 N$. The string is set into vibration using an external vibrator of frequency $100 Hz$. Find the separation (in $cm$ ) between the successive nodes on the string.
(2009)
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Answer:
Correct Answer: 40. (5)
Solution:
- Distance between the successive nodes,
$$ d=\frac{\lambda}{2}=\frac{v}{2 f}=\frac{\sqrt{T / \mu}}{2 f} $$
Substituting the values we get, $d=5 cm$
