Home
Class 12
PHYSICS
Two waves passing through a region are r...

Two waves passing through a region are represented by `y=(1.0cm) sin [(3.14 cm^(-1))x - (157s^(-1))t]`
and `y = (1.5 cm) sin [(1.57 cm^(-1))x- (314 s^(-1))t].`
Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.

Text Solution

Verified by Experts

Accoding to the principle of superposition, each wave produce its disturbance independent of the other and the resultant disturbance is equal to the vector sum of the individual disturbance. The displacements of the particle at `x = 1cm` at time `t = 5.0 ms` due to the two waves are.
`y_(1) = 2 mm [(2pi cm^(-1)) xx - (50 pi s^(-1))t]`
`y_(1) = 5 mm sin[(2pi cm^(-1)) xx 1 cm - (50 pi s^(-1))5 xx 10^(-3) sec]`
`= 5 mm sin [2pi - (p)/(4)] = -5 mm`
and `y_(2) = 10 mm sin [(pi cm^(-1)) xx - (100 pi s^(-1))t]`,
`y_(2) = 10 mm sin [(pi cm^(-1)) xx 1 cm - (100 pi s^(-1))5 xx 10^(-3) sec]`
`= 10 mm sin [pi - (pi)/(2)] = 10 mm`
The net displacement is : `y = y_(1) + y_(2) = 10 mm - 5 mm = 5 mm`
Promotional Banner

Topper's Solved these Questions

  • TRAVELLING WAVES

    RESONANCE ENGLISH|Exercise Solved Miscellaneous Problems|7 Videos
  • TRAVELLING WAVES

    RESONANCE ENGLISH|Exercise Board Level Exercise|27 Videos
  • TEST SERIES

    RESONANCE ENGLISH|Exercise PHYSICS|130 Videos
  • WAVE ON STRING

    RESONANCE ENGLISH|Exercise Exercise- 3 PART II|7 Videos

Similar Questions

Explore conceptually related problems

Two waves passing through a region are represented by y=(1.0cm) sin [(3.14 cm^(-1))x - (157s^(-1) x - (157s^(-1))t] and y = (1.5 cm) sin [(1.57 cm^(-1))x- (314 s^(-1))t]. Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.

Two waves passing through a region are represented by y = ( 1.0 m) sin [( pi cm^(-1)) x - ( 50 pi s^(-1))t] and y = ( 1.5 cm) sin [( pi //2 cm^(-1)) x - ( 100 pi s^(-1)) t] . Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms .

The equation for a wave travelling in x-direction 0n a string is y = (3.0 cm) sin [(3.14 cm^(-1) x -(314 s^(-1)t] (a) Find the maximum velocity of a particle of the string. (b) Find the acceleration of a particle at x = 6.0 cm at time t = 0.11 s

The equation for a wave travelling in x-direction on a string is y = (3.0cm)sin[(3.14 cm^(-1) x - (314s^(-1))t] (a) Find the maximum velocity of a particle of the string. (b) Find the acceleration of a particle at x =6.0 cm at time t = 0.11 s.

The equation of a travelling wave on a string is y = (0.10 mm ) sin [31.4 m^(-1) x + (314s^(-1)) t]

The vibrations of a string fixed at both ends are described by the equation y= (5.00 mm) sin [(1.57cm^(-1))x] sin [(314 s^(-1))t] (a) What is the maximum displacement of particle at x = 5.66 cm ? (b) What are the wavelengths and the wave speeds of the two transvers waves that combine to give the above vibration ? (c ) What is the velocity of the particle at x = 5.66 cm at time t = 2.00s ? (d) If the length of the string is 10.0 cm, locate the nodes and teh antinodes. How many loops are formed in the vibration ?

A wave is represented by the equation y=(0.001mm)sin[(50s^-1t+(2.0m^-1)x]

Two waves travelling in a medium in the x-direction are represented by y_(1) = A sin (alpha t - beta x) and y_(2) = A cos (beta x + alpha t - (pi)/(4)) , where y_(1) and y_(2) are the displacements of the particles of the medium t is time and alpha and beta constants. The two have different :-

Two waves are represented by: y_(1)=4sin404 pit and y_(2)=3sin400 pit . Then :

The equation of a wave travelling on a string is given by Y(mn) = 8 sin[ (5m^(-1)x-(4s^(-1)t ]. Then

RESONANCE ENGLISH-TRAVELLING WAVES-Exercise- 3 PART I
  1. Two waves passing through a region are represented by y=(1.0cm) sin [(...

    Text Solution

    |

  2. A harmonically moving transverse wave on a string has a maximum partic...

    Text Solution

    |

  3. A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends...

    Text Solution

    |

  4. When two progressive waves y(1)=4sin(2x-6t)andy(2)=3sin(2x-6t-(pi)/(2)...

    Text Solution

    |

  5. A massless rod of length l is hung from the ceiling with the help of t...

    Text Solution

    |

  6. A transverse sinusoidal wave moves along a string in the positive x-di...

    Text Solution

    |

  7. A horizontal stretched string, fixed at two ends, is vibrating in its ...

    Text Solution

    |

  8. One end of a taut string of length 3 m along the x-axis is fixed at x ...

    Text Solution

    |

  9. A string is stretched betweeb fixed points separated by 75.0 cm. It ob...

    Text Solution

    |

  10. A steel wire of length 50sqrt(3) cm is connected to an aluminium wire ...

    Text Solution

    |

  11. An aluminium wire of cross-sectional area (10^-6)m^2 is joined to a st...

    Text Solution

    |

  12. The fundatmental frequency of a sonometer wire increases by 6 Hz if it...

    Text Solution

    |

  13. A metal wire with volume density rho and young's modulus Y is stretche...

    Text Solution

    |

  14. Find velocity of wave is string A &B.

    Text Solution

    |

  15. A string of length 50 cm is vibrating with a fundamental frequency of ...

    Text Solution

    |

  16. A string fixed at both is vibrating in the lowest mode of vibration fo...

    Text Solution

    |

  17. A guitar string is 90 cm long and has a fundamental frequency of 124 H...

    Text Solution

    |

  18. A piano wire weighing 6.00 g and having a length of 90.0 cm emits a fu...

    Text Solution

    |

  19. Length of a sonometer wire is 1.21 m. Find the length of the three seg...

    Text Solution

    |

  20. In the figure shown A and B are two ends of a string of length 100m. S...

    Text Solution

    |