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 \[=512\text{ }Hz\,\,x=\] 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, \[{{n}_{p}}\uparrow -{{n}_{f}}=x\downarrow \xrightarrow[{}]{{}}worng\] |
| \[{{n}_{p}}-{{n}_{p}}\uparrow =x\downarrow \xrightarrow[{}]{{}}correct\] |
| \[\Rightarrow \] \[{{n}_{p}}={{n}_{f}}-x=512-4=508\,Hz\] |
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