Expression for Standing Waves
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Q. Two identical transverse sinusoidal waves travel in opposite directions along a string. The speed of transverse waves in the string is 0.5 cm/s. Each has an amplitude of 3.0 cm and wavelength of 6.0 cm. The equation for the resultant wave is
- y=6sin(πt6)cos(πx3)
- y=6sin(πx3)cos(πt6)
- Both (a) and (b) may be correct
- Both (a) and (b) are wrong
Q. Vibration in a string of length 60 cm fixed at both ends is represented by the equation, y=4sin(πx15)cos(96πt), where x and y are in centimeter and t in seconds. The number of loops formed in vibrating string will be
- 4
- 3
- 6
- 5
Q. A string with a mass density of 4×10−3 kg/m is under a tension of 360 N and is fixed at both ends. One of its resonance frequencies is 375 Hz. The next higher resonance frequency is 450 Hz. If the mass of the string is represented by x×10−3 kg, find x
Q. When a stationary wave is formed, then its frequency is that of the individual waves.
- same as
- twice
- half
Q. The equation of a standing wave in a stretched string is given by y=5sin(πx3)cos(40πt), where x and y are in cm and t is in sec. The separation between two consecutive nodes is (in cm)
- 1.5
- 3
- 6
- 4
Q. For a string clamped at both ends, which of the following wave equation is valid for a stationary wave set up in it whose origin is at one of the ends of the string?
- y=Asinkxsinωt
- y=Acoskxsinωt
- y=Acoskxcosωt
- y=2Acoskxcosωt
Q.
The vibrations of a string fixed at both ends are described by the
y=(5.00mm)sin[(1.57cm−1)x]sin[(314s−1)t]
If the length of the string is 10.0 cm, locate the nodes and the antinodes. How many loops are formed in the vibration?
nodes - 0, 2cm, 4 cm, 6 cm, 8 cm, 10 cm
Anti-nodes - 1 cm, 3 cm, 5 cm, 7 cm, 9 cm
nodes - 0, 1 cm, 3 cm, 9 cm
Anti-nodes - 2 cm, 4 cm, 6 cm, 8 cm, 10 cm
nodes - 0, 4 cm, 8 cm, 10 cm
Anti-nodes - 1 cm, 5 cm, 9 cm
None of these
Q. A string of length l along x− axis is fixed at both ends and is vibrating in second harmonic. If amplitude of incident wave is 2.5 mm, the equation of standing wave is (T is tension and μ is linear density)
- (2.5mm) sin (2πlx)cos (2π√(Tμl2)t)
- (5mm) sin (πlx)cos 2πt
- (5mm) sin (2πlx)cos (2π√(Tμl2)t)
- (7.5mm) cos (2πlx)cos (2π√(Tμl2)t)
Q. A plane simple harmonic progressive wave given by the equation, y=Asin(ωt−kx) of wavelength 120 cm is incident normally on a plane surface which is a perfect reflector (acts as fixed end). If a stationary wave is formed, then the ratio of amplitudes of vibrations at points 10 cm and 30 cm from the reflector is
(Here x is measured from reflector)
(Here x is measured from reflector)
- 1:2
- 1:3
- 2:1
- 3:1
Q. If in a stationary wave the amplitude corresponding to antinode is 4 cm, then the amplitude corresponding to a particle of medium located exactly midway between a node and an antinode is
- 2 cm
- 2√2 cm
- √2 cm
- 1.5 cm
Q. A standing wave y=A sin (20πx3)cos (1000 π t) is set up in a taut string, where x and y are in metre and t is in seconds. The distance between first two successive points oscillating with the amplitude A2, from the fixed end will be equal to
- 10 cm
- 15 cm
- 2.5 cm
- 4 cm
Q. A standing wave y=A sin (20πx3)cos (1000 π t) is set up in a taut string, where x and y are in metre and t is in seconds. The distance between first two successive points oscillating with the amplitude A2, from the fixed end will be equal to
- 10 cm
- 15 cm
- 2.5 cm
- 4 cm
Q. A standing wave y=A sin (20πx3)cos (1000 π t) is set up in a taut string, where x and y are in metre and t is in seconds. The distance between first two successive points oscillating with the amplitude A2, from the fixed end will be equal to
- 10 cm
- 15 cm
- 2.5 cm
- 4 cm
Q. Spacing between two successive nodes in a standing wave on a string is x. If tension in the string is doubled keeping frequency constant, then find the new spacing between successive nodes.
- x√2
- √2x
- x2
- 2x
Q. Spacing between two successive nodes in a standing wave on a string is x. If tension in the string is doubled keeping frequency constant, then find the new spacing between successive nodes.
- x√2
- √2x
- x2
- 2x
Q. Equation of a stationary wave is given by y=5cos(πx25)sin(100πt). Here, x is in centimeter and t in seconds. Node will not occur at distance
- 25 cm
- 62.5 cm
- 12.5 cm
- 37.5 cm