A) \[\frac{8{{f}_{0}}\Delta \ell }{\ell }\]
B) \[\frac{{{f}_{0}}\Delta \ell }{\ell }\]
C) \[\frac{2{{f}_{0}}\Delta \ell }{\ell }\]
D) \[\frac{4{{f}_{0}}\Delta \ell }{\ell }\]
Correct Answer: A
Solution :
| \[{{f}_{0}}=\frac{\text{v}}{2\ell }\] |
| Now beat frequency \[=({{f}_{1}}-{{f}_{2}})\] |
| \[=\frac{\text{v}}{2\left( \frac{\ell }{2}-\Delta \ell \right)}-\frac{\text{v}}{2\left( \frac{\ell }{2}+\Delta \ell \right)}=\frac{\text{v}}{2}\left[ \frac{1}{\frac{\ell }{2}-\Delta \ell }-\frac{1}{\frac{\ell }{2}+\Delta \ell } \right]\] |
| \[=({{f}_{0}}\ell )\left[ \frac{2}{\ell -2\Delta \ell }-\frac{2}{\ell +2\Delta \ell } \right]\] |
| \[=2{{f}_{0}}\ell \left[ \frac{\ell +2\Delta \ell -\ell +2\Delta \ell }{{{\ell }^{2}}-4{{(\Delta \ell )}^{2}}} \right]\approx 2{{f}_{0}}\ell \left( \frac{4\Delta \ell }{{{\ell }^{2}}} \right)\approx \frac{8{{f}_{0}}\Delta \ell }{\ell }\] |
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