JEE Physics 2024 Question with Solution

Given: A closed organ pipe has length $150 \, \text{cm} = 1.5 \, \text{m}$ and an open organ pipe has length $350 \, \text{cm} = 3.5 \, \text{m}$. Both vibrate in fundamental mode and the beat frequency is $7 \, \text{Hz}$.

Find: The velocity of sound $v$.

For a closed organ pipe in fundamental mode,

$$f_c = \frac{v}{4L_c}$$

So,

$$f_c = \frac{v}{4 \times 1.5} = \frac{v}{6}$$

For an open organ pipe in fundamental mode,

$$f_o = \frac{v}{2L_o}$$

So,

$$f_o = \frac{v}{2 \times 3.5} = \frac{v}{7}$$

Using beat frequency,

$$|f_o - f_c| = 7$$

Substituting the two frequencies,

$$\left|\frac{v}{7} - \frac{v}{6}\right| = 7$$ $$\left|\frac{6v - 7v}{42}\right| = 7$$ $$\left|\frac{-v}{42}\right| = 7$$ $$\frac{v}{42} = 7$$ $$v = 42 \times 7 = 294 \, \text{m/s}$$

Therefore, the velocity of sound is $294 \, \text{m/s}$.

Given: The closed pipe length is $1.5 \, \text{m}$ and the open pipe length is $3.5 \, \text{m}$. The beat frequency is $7 \, \text{Hz}$.

Find: The value of $v$.

The fundamental frequency of the closed pipe is

$$f_c = \frac{v}{4L} = \frac{v}{4 \times 1.5} = \frac{v}{6}$$

and the fundamental frequency of the open pipe is

$$f_o = \frac{v}{2L} = \frac{v}{2 \times 3.5} = \frac{v}{7}$$

Now the beat frequency is the absolute difference of these frequencies:

$$|f_c - f_o| = 7$$ $$\left|\frac{v}{6} - \frac{v}{7}\right| = 7$$

Taking the difference,

$$\left|\frac{7v - 6v}{42}\right| = 7$$ $$\frac{v}{42} = 7$$ $$v = 294$$

Thus, the required velocity of sound is $294 \, \text{m/s}$.