A) 196 Hz
B) 284 Hz
C) 375 Hz
D) 460 Hz
Correct Answer: A
Solution :
The frequency, when a sonometer wire of vibrating length is 48 cm. \[{{v}_{1}}=\frac{c}{2\times {{l}_{1}}}=\frac{c}{2\times 0.48}=\frac{c}{0.96}\] The frequency when a sonometer wire of vibrating length is 50 cm \[{{v}_{2}}=\frac{c}{2\times {{l}_{2}}}=\frac{c}{2\times 0.50}=\frac{c}{1.00}\] \[\Rightarrow \,{{v}_{1}}-{{v}_{2}}=8\Rightarrow {{v}_{1}}+{{v}_{2}}+8\] \[\Rightarrow \,\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1.00}{0.96}\Rightarrow {{v}_{1}}=\frac{1.00}{0.96}{{v}_{2}}\] \[\therefore \,\,{{v}_{2}}+8\,={{v}_{2}}\times \,\frac{100}{96}\,\Rightarrow \,{{v}_{2}}=196\,Hz\] The frequency of the tuning fork \[v={{v}_{2}}+4=192+4=196\,Hz\].You need to login to perform this action.
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