Wave Mechanics Question 653
Question: The equation of a wave on a string of linear mass density $ 0.04kg,{{m}^{-1}} $ is given by $ y=0.02(m)\sin [ 2\pi ( \frac{1}{0.04(s)}-\frac{x}{0.50(m)} ) ] $ The tension in the string is
Options:
A) 4.0 N
B) 12.5 N
C) 0.5 N
D) 6.25 N
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ T=\mu ,v^{2}=\mu \frac{{{\omega }^{2}}}{k^{2}}=0.04\frac{{{(2\pi /0.004)}^{2}}}{{{(2\pi /0.50)}^{2}}} $ 6.25 N
BETA
