Problem Analysis: The question involves calculating the frequency of a tuning fork based on the number of beats heard when sounded with a sonometer wire at two different lengths. The key concepts are the phenomenon of beats and the relationship between the frequency of a sonometer wire and its length.

Understanding Beats and Sonometer Frequency

Beats occur when two sound waves of slightly different frequencies interfere. The number of beats per second is equal to the absolute difference between the two frequencies. For a sonometer wire, the fundamental frequency ($f$) is inversely proportional to its length ($L$), assuming tension ($T$) and linear density ($m$) are constant. Mathematically, $f \propto 1/L$.

Setting Up the Equations

Let $f_f$ be the frequency of the tuning fork. Let $f_{w1}$ be the frequency of the sonometer wire when its length is $L_1 = 0.95$ m. Let $f_{w2}$ be the frequency of the sonometer wire when its length is $L_2 = 1$ m.

Since frequency is inversely proportional to length, a shorter length produces a higher frequency. Thus, $f_{w1} > f_{w2}$.

We are given that 5 beats/second are heard in both cases. This means:

1. $|f_f - f_{w1}| = 5$
2. $|f_f - f_{w2}| = 5$

From these two equations, we can infer that the tuning fork's frequency $f_f$ must lie exactly midway between the two wire frequencies $f_{w1}$ and $f_{w2}$. This implies:

$f_f = \frac{f_{w1} + f_{w2}}{2}$

Also, the difference between the wire frequencies is twice the beat frequency:

$f_{w1} - f_{w2} = 2 \times 5 = 10$ Hz

Calculating Wire Frequencies

We know $f_w = k/L$, where $k$ is a constant related to tension and linear density.

$f_{w1} = k / 0.95$

$f_{w2} = k / 1$

Substitute these into the difference equation:

$\frac{k}{0.95} - \frac{k}{1} = 10$

$k \left( \frac{1}{0.95} - 1 \right) = 10$

$k \left( \frac{1 - 0.95}{0.95} \right) = 10$

$k \left( \frac{0.05}{0.95} \right) = 10$

$k = 10 \times \frac{0.95}{0.05} = 10 \times 19 = 190$

Now, calculate $f_{w1}$ and $f_{w2}$:

$f_{w1} = \frac{190}{0.95} = 200$ Hz

$f_{w2} = \frac{190}{1} = 190$ Hz

Check: $f_{w1} - f_{w2} = 200 - 190 = 10$ Hz. This is consistent.

Determining Tuning Fork Frequency

Use the formula derived earlier:

$f_f = \frac{f_{w1} + f_{w2}}{2}$

$f_f = \frac{200 \text{ Hz} + 190 \text{ Hz}}{2}$

$f_f = \frac{390 \text{ Hz}}{2}$

$f_f = 195$ Hz

Verify the beats:

$|f_f - f_{w1}| = |195 - 200| = |-5| = 5$ beats/sec

$|f_f - f_{w2}| = |195 - 190| = |5| = 5$ beats/sec

Both conditions are satisfied.