Two waves are represented by the equations y_{1} = a sin (ωt + kx + 0.57) m and y_{2} = a cos (ωt + kx) m, where x is in meter and t in sec. The phase difference between them is [2011]
Here, y_{1} = a sin (ωt + kx + 0.57)
and y_{2} = a cos (ωt + kx)
Phase difference,
= 1.57 – 0.57
= 1 radian
Sound waves travel at 350 m/s through a warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air [2011]
Two identical piano wires kept under the same tension T have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate together would be [2011M]
For fundamental mode,
Taking logarithm on both sides, we get
Differentiating both sides, we get
(as and μ are constants)
Here df = 6
f = 600 Hz
When a string is divided into three segments of length l_{1}, l_{2}, and l_{3} the fundamental frequencies of these three segments are v_{1}, v_{2} and v_{3} respectively. The original fundamental frequency (v) of the string is [2012]
Answer : c
Solution : Let l be the length of the string
Frequency fundamental is given by (v) = 1/2l(T/u)^½
V is inversely proportional to the l
Here, l1 = k/v1 ; l2 = k/v2; l3 = k/v3; l =k/v
But l = l1 + l2 + l3
1/v = 1/v1 + 1/v2 + 1/v3
Two sources of sound placed close to each other are emitting progressive waves given by y_{1} = 4 sin 600πt and y_{2} = 5 sin 608πt. An observer located near these two sources of sound will hear :[2012]
2πf_{1} = 600π
f_{1} = 300 ... (1)
2πf_{2} = 608π
f_{2} = 304 ...(2)
f_{1} – f_{2} = 4 beats
where A_{1}, A_{2 }are amplitudes of given two sound wave.
A train moving at a speed of 220 ms^{–1} towards a stationary object, emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is : [2012M] (speed of sound in air is 330 ms^{–1})
Frequency of the echo detected by the driver of the train is (According to Doppler effect in sound)
where f = original frequency of source of sound f' = Apparent frequency of source because of the relative motion between source and observer.
If we study the vibration of a pipe open at both ends, then which of the following statements is not true ? [NEET 2013]
Pressure change will be minimum at both ends. In fact, pressure variation is maximum at /2 because the displacement node is pressure antinode.
A source of unknown frequency gives 4 beats/s, when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency 513 Hz. The unknown frequency is [NEET 2013]
When sounded with a source of known frequency fundamental frequency = 250 ± 4 Hz = 254 Hz or 246 Hz
2^{nd} harmonic if unknown frequency (suppose) 254 Hz = 2 × 254 = 508 Hz
As it gives 5 beats
∴ 508 + 5 = 513 Hz
Hence, unknown frequency is 254 Hz
A wave travelling in the +ve xdirection having displacement along ydirection as 1m, wavelen gth 2πm and frequency Hz isrepresented by [NEET 2013]
As Y = A sin (ωt – kx + φ)
The length of the wire between two ends of a sonometer is 100 cm. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio of 1 : 3 : 5? [NEET Kar. 2013]
Answer : a
Solution : Fundamental frequency (v) = 1/L
As the fundamental frequencies are in the ratio of 1:3:5,
L1 : L2 : L3 = 1/1 : 1/3 : 1/5 => 15 : 5 : 3
Let x be the common factor , then
15x + 5x + 3x = L = 100
23x = 100
=> x = 100/23
L1 = 15*100/23 = 1500/23
L2 = 5*100/23 = 500/23
L3 = 3*100/23 = 300/23
The bridge should be placed from A at (1500/23 , 2000/23)
Two sources P and Q produce notes of frequency 660 Hz each. A listener moves from P to Q with a speed of 1 ms^{–1}. If the speed of sound is 330 m/s, then the number of beats heard by the listener per second will be [NEET Kar. 2013]
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