Advertisements
Advertisements
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?
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.
APPEARS IN
संबंधित प्रश्न
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2 : 1 the radii are in the ratio 3 : 1 and the densities are in the ratio 1 : 2. Find the ratio of their fundamental frequencies.
Two wires of same material are vibrating under the same tension. If the first overtone of first wire is equal to the second overtone of second wire and radius of first wire is twice the radius of the second then the ratio of length of first wire to second wire is
The number of possible natural oscillations of the air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz? (v = 340 m/s)
Water waves produced by a motor boat sailing in water are ______.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz? (sound velocity in air = 330 ms–1)
A wire of length 2L is made by joining two wires A and B of the same length but different radii r and 2r, and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is ______.
Two closed end pipes when sounded together produce 5 beat per second. If their length are in the ratio 100 : 101, then fundamental notes produced by them are ______.
A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ______ cm.
(Take the speed of sound in air as 340 ms-1.)
A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is ______.
