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Waves · Question 52

AIPMT 2002MCQEasy📈 Topic in 25% papers+4−1
1

A wave travelling in positive X-direction with a = 0.2 m, velocity = 360 m s-1 and = 60 m, then correct expression for the wave is

  1. y = 0.2 [2 (6t + x60 ) ]
  2. y = 0.2 [ (6t + x60 ) ]
  3. y = 0.2 [2 (6t - x60 ) ]
  4. y = 0.2 [ (6t - x60 ) ]

Answer

(3) y = 0.2 [2 (6t - x60 ) ]

Solution

Frequency: = v = 36060 = 6 Hz

A wave along +x: y = a [2 ( t - x ) ]

y = 0.2 [2 (6t - x60 ) ]

Answer: (3)

AIPMT 2002

Keep practising — Waves

2

A whistle revolves in a circle with angular speed = 20 rad/s using a string of length 50 cm. If the frequency of sound from the whistle is 385 Hz, then what is the minimum frequency heard by an observer which is far away from the centre (velocity of sound = 340 m/s)

  1. 385 Hz
  2. 374 Hz
  3. 394 Hz
  4. 333 Hz
Out of syllabusAIPMT 2002
3

The equation of a wave is represented by y = 10-4 (100t - x10 ) m, then the velocity of wave will be

  1. 100 m/s
  2. 4 m/s
  3. 1000 m/s
  4. 10 m/s
AIPMT 2001
4

Two waves having equation x1 = a ( t - kx + 1), x2 = a ( t - kx + 2). If in the resultant wave the frequency and amplitude remain equal to amplitude of superimposing waves, the phase difference between them is

  1. 6
  2. 2 3
  3. 4
  4. 3
AIPMT 2001
5

If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved then its fundamental frequency will become

  1. n4
  2. 2n
  3. n
  4. n 2
AIPMT 2001
6

A string is cut into three parts, having fundamental frequencies n1, n2, n3 respectively. Then original fundamental frequency n related by the expression as

  1. 1n = 1n1 + 1n2 + 1n3
  2. n = n1 n2 n3
  3. n = n1 + n2 + n3
  4. n = n1 + n2 + n33
AIPMT 2000
7

Two stationary sources each emit waves of wavelength . An observer moves from one source to the other with velocity u. Then the number of beats heard by him is

  1. 2u
  2. u
  3. u
  4. u2
Out of syllabusAIPMT 2000

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