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प्रश्न
A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water (Figure). The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, maximum intensity of sound is heard. If the room temperature is 20°C, calculate
- speed of sound in air at room temperature
- speed of sound in air at 0°C
- if the water in the tube is replaced with mercury, will there be any difference in your observations?
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उत्तर
a. Pipe partially filled with water behaves like a one-end open organ pipe. In the 1st harmonic, one antinode and one node at water level are formed, so the first harmonic or at maximum intensity is heard at L = 17 cm. So,
L = `λ/4`
⇒ λ = 4L
⇒ λ = 4 × 17
⇒ λ = 68 cm
⇒ λ = 0.68 cm
Frequency of tuning fork, f = 512 Hz
The velocity of sound in air, v = λf. Therefore, we get
⇒ v = 0.68 × 512 m/s
⇒ v = 348.16 m/s
Thus, the velocity of sound in air at room temperature 20°C is 348.16 m/s
b. We know that `vαsqrt(T)`
⇒ `v_o/v_T = sqrt(T_0/T)`
T0 = 273 K
vT = v20 = 348.16
T = 20 + 273 = 293 K
⇒ `v_0 = v_T sqrt(T_0/T_T)`
⇒ `v_0 = 348.16 sqrt(273/293)`
⇒ `v_0 = 348.16 sqrt(0.9317)`
⇒ v0 = 348.16 × 0.96526
⇒ v0 = 336 m/s
c. The sound is reflected into the air column by water and mercury to form a stationary wave. Mercury is denser than water, this reflection is more in mercury than water. So, the intensity of sound heard will be larger but the reading doesn’t change as the medium in a tube (air) and tuning fork are the same.
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