wave_mechanics critical_thinking Question 10
Question: Two identical stringed instruments have frequency 100 Hz. If tension in one of them is increased by 4% and they are sounded together then the number of beats in one second is
[EAMCET (Engg.) 1995]
Options:
A) 1
B) 8
C) 4
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
Frequency of vibration in tight string $ n=\frac{p}{2l}\sqrt{\frac{T}{m}}\Rightarrow n\propto \sqrt{T} $
Therefore $ \frac{\Delta n}{n}=\frac{\Delta T}{2T}=\frac{1}{2}\times (4%)=2% $
Therefore Number of beats = $ \Delta n=\frac{2}{100}\times n=\frac{2}{100}\times 100=2 $
BETA
