A) 510 Hz
B) 514-Hz
C) 516 Hz
D) 508 Hz
Correct Answer: D
Solution :
Suppose \[{{n}_{p}}=\] frequency of piano =? \[({{n}_{p}}\propto \sqrt{T})\]\[{{n}_{f}}=\]frequency of tuning fork = 512 Hz x = Beat frequency = 4 beats/s, which is decreasing \[(4\to 2)\] after changing the tension of piano wire. Also, tension of piano wire is increasing so \[{{n}_{p}}\uparrow \] Hence, \[{{n}_{p}}\uparrow -{{n}_{f}}=x\downarrow \xrightarrow[{}]{{}}\text{worng}\] \[{{n}_{p}}-{{n}_{p}}\uparrow =x\downarrow \xrightarrow[{}]{{}}\text{correct}\] \[\Rightarrow {{n}_{p}}={{n}_{f}}-x=512-4=508\,Hz\]
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