Question

Wire having tension $225\, N$ produces six beats per second when it is tuned with a fork. When tension changes to $256\, N$, it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be

• A. $186 \,Hz$
• B. $225 \,Hz$
• C. $256 \,Hz$
• D. $280 \,Hz$

Answer 1

Answer: (A)

We know that, for a string, frequency is proportional to square root of Tension in the string.
i.e., $f \propto \sqrt{ T }$
Let the tuning fork frequency be $f$ and frequency of the string be $f_{1}$ and $f_{2}$ for the values of tension as $225 N$ and $256\, N$ respectively.
Thus, $\frac{ f _{1}}{ f _{2}}=\sqrt{\frac{225}{256}}=\frac{15}{16}$
As the tuning fork produces $6$ beats per second on each of the case,
we have $f - f _{1}=6$ and $f _{2}- f =6$
Using $16 f _{1}=15 f _{2}$,
we have $15\left( f _{2}- f \right)+16\left( f - f _{1}\right)=(16+15) \times 6$ $\Rightarrow f =31 \times 6=186\, Hz$