A) 200 cm
B) 230 cm
C) 210 cm
D) 250 cm
Correct Answer: D
Solution :
\[Y=A\sin \frac{2\pi }{\lambda }=(vt-x)\] \[\frac{Y}{A}=\sin 2\pi \left( \frac{t}{T}-\frac{x}{\lambda } \right)\] In first case, \[\frac{{{Y}_{1}}}{A}=\sin 2\pi \left( \frac{t}{T}-\frac{{{x}_{1}}}{\lambda } \right)\] Here, \[{{Y}_{1}}=+6,\,\,A=8,\,\,{{x}_{1}}=10\,cm\] \[\frac{6}{8}=\sin 2\pi \left( \frac{t}{T}-\frac{25}{\lambda } \right)\] ... (i) Similarly in the second case, \[\frac{4}{8}=\sin 2\pi \left( \frac{t}{T}-\frac{25}{\lambda } \right)\] ... (ii) From Eq. (i), \[2\lambda \left( \frac{t}{T}-\frac{10}{\lambda } \right)={{\sin }^{-1}}\left( \frac{6}{8} \right)=0.85\]rad \[\Rightarrow \] \[\frac{t}{T}-\frac{25}{\lambda }=0.08\] ? (iii) Similarly from Eq. (ii), \[2\pi \left( \frac{t}{T}-\frac{25}{\lambda } \right)={{\sin }^{-1}}\left( \frac{4}{8} \right)=\frac{\pi }{6}\] rad \[\Rightarrow \] \[\frac{t}{T}-\frac{25}{\lambda }=0.08\] ... (iv) Subtracting Eq. (iv) from Eq. (iii), we get \[\frac{15}{\lambda }=0.06\Rightarrow \lambda =250\,cm\]
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