question_answer

Two strings X and Y of a sitar produce a beat frequency 4 Hz. When the tension of the string Y is slightly increased the beat frequency is found to be 2 Hz. If the frequency of X is 300 Hz, then the original frequency of Y was  [UPSEAT 2000]

A)            296 Hz                                     

B)            298 Hz

C)            302 Hz                                     

D)            304 Hz

Correct Answer: A

Solution :

           \[{{n}_{x}}=300Hz,\] \[{{n}_{y}}=?\] x = beat frequency = 4 Hz, which is decreasing (4®2) after increasing the tension of the string y. Also tension of wire y increasing so \[{{n}_{y}}\uparrow \] \[(\because \,n\propto \sqrt{T})\] Hence \[{{n}_{x}}-{{n}_{y}}\downarrow =x\downarrow \]             Correct             \[{{n}_{y}}\uparrow -{{n}_{x}}=x\downarrow \]                     Wrong Þ \[{{n}_{y}}={{n}_{x}}-x=300-4=296Hz\]