A) \[\text{2}00\text{cm}\]
B) \[\text{23}0\text{ cm}\]
C) \[\text{21}0\text{ cm}\]
D) \[\text{25}0\text{ cm}\]
Correct Answer: D
Solution :
\[\text{Y=A sin}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{ }\!\!\lambda\!\!\text{ }}\text{ (vt-x)}\] \[\frac{Y}{A}=\sin 2\pi \left( \frac{t}{T}-\frac{x}{\lambda } \right)\] In first case, \[\frac{{{Y}_{1}}}{A}=\sin 2\pi \left( \frac{t}{T}-\frac{{{x}_{1}}}{\lambda } \right)\] Here, \[{{Y}_{1}}=\text{ }+\text{ }6,A\text{ }=\text{ }8,\text{ }{{x}_{1}}=10\text{ }cm\] \[\frac{6}{8}=\sin 2\pi \left( \frac{t}{T}-\frac{10}{\lambda } \right)\] ? (i) Similarly in the second case, \[\frac{4}{8}=\sin 2\pi \left( \frac{t}{T}-\frac{25}{\lambda } \right)\] ? (ii) From Eq. (i), \[2\lambda \left( \frac{t}{T}-\frac{10}{\lambda } \right)={{\sin }^{-1}}\left( \frac{6}{8} \right)=\text{0}\text{.85 rad}\] \[\Rightarrow \] \[\frac{t}{T}-\frac{10}{\lambda }=0.14\] ? (iii) Similarly form Eq. (ii), \[2\pi \left( \frac{t}{T}-\frac{25}{\lambda } \right)={{\sin }^{-1}}\left( \frac{4}{8} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{ rad}\] \[\Rightarrow \] \[\frac{t}{T}-\frac{25}{\lambda }=0.08\] ? (iv) Subtracting Eq. (iv) from Eq. (iii), we get \[\frac{15}{\lambda }=0.06\] \[\Rightarrow \] \[\text{ }\!\!\lambda\!\!\text{ =250 cm}\]
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