A) 0.12 N
B) 0.21 N
C) 1.2 N
D) 0.012 N
Correct Answer: A
Solution :
Mass per unit length\[m=1.3\times {{10}^{-4}}kg/m\] \[y=0.021\sin (x+30t)\] Comparing with\[y=a\sin \left( \omega t=\frac{2\pi x}{\lambda } \right)\] \[a=0.021,\omega =30\] \[\Rightarrow \] \[n=\frac{30}{2\pi }\] \[\frac{2\pi }{\lambda }=1\] \[\lambda =\frac{\lambda }{2\pi }\] Frequency\[n=\frac{1}{\lambda }\frac{\sqrt{1}}{m}\] \[\frac{30}{2\pi }=\frac{1}{2\pi }\sqrt{\frac{T}{1.3\times {{10}^{-4}}}}\] \[T=1.3\times {{10}^{-4}}\times 900\] \[=0.117=0.12N\]You need to login to perform this action.
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