Home
Class 11
PHYSICS
A sound wave with an amplitude of 3cm st...

A sound wave with an amplitude of `3cm` starts towards right from origin and gets reflected at a rigid wall after a second. If the velocity of the wave is `340ms^(-1)` and it has a wavelength of `2m`, the equations of incident and reflected waves respectively are :

A

`y=3xx10^(-2)sinpi(340t-x)`, `y=-3xx10^(-2)sinpi(340t+x)` towards left

B

`y=3xx10^(-2)sinpi(340t+x)`, `y=-3xx10^(-2)sinpi(340t+x)` towards left

C

`y=3xx10^(-2)sinpi(340t-x)`, `y=-3xx10^(-2)sinpi(340t-x)` towards left

D

`y=3xx10^(-2)sinpi(340t-x)`, `y=3xx10^(-2)sinpi(340t+x)` towards left

Text Solution

AI Generated Solution

The correct Answer is:
To find the equations of the incident and reflected sound waves, we can follow these steps: ### Step 1: Identify Given Information - Amplitude (A) = 3 cm = 3 × 10^(-2) m - Velocity (v) = 340 m/s - Wavelength (λ) = 2 m ### Step 2: Calculate Wave Number (k) The wave number (k) is given by the formula: \[ k = \frac{2\pi}{\lambda} \] Substituting the value of λ: \[ k = \frac{2\pi}{2} = \pi \, \text{(rad/m)} \] ### Step 3: Calculate Angular Frequency (ω) The angular frequency (ω) can be calculated using the relationship between velocity, wavelength, and frequency: \[ v = f \lambda \] From this, we can express frequency (f) as: \[ f = \frac{v}{\lambda} = \frac{340}{2} = 170 \, \text{Hz} \] Now, we can find ω: \[ \omega = 2\pi f = 2\pi \times 170 = 340\pi \, \text{(rad/s)} \] ### Step 4: Write the Equation of the Incident Wave The standard form of the wave equation is: \[ y = A \sin(\omega t - kx) \] Substituting the values of A, ω, and k: \[ y_{\text{incident}} = 3 \times 10^{-2} \sin(340\pi t - \pi x) \] ### Step 5: Write the Equation of the Reflected Wave When a wave reflects off a rigid wall, it undergoes a phase shift of π. Therefore, the equation of the reflected wave will be: \[ y_{\text{reflected}} = A \sin(\omega t + kx + \pi) \] Using the identity \( \sin(x + \pi) = -\sin(x) \): \[ y_{\text{reflected}} = -3 \times 10^{-2} \sin(340\pi t + \pi x) \] ### Final Equations Thus, the equations of the incident and reflected waves are: 1. **Incident Wave**: \( y_{\text{incident}} = 3 \times 10^{-2} \sin(340\pi t - \pi x) \) 2. **Reflected Wave**: \( y_{\text{reflected}} = -3 \times 10^{-2} \sin(340\pi t + \pi x) \)

To find the equations of the incident and reflected sound waves, we can follow these steps: ### Step 1: Identify Given Information - Amplitude (A) = 3 cm = 3 × 10^(-2) m - Velocity (v) = 340 m/s - Wavelength (λ) = 2 m ### Step 2: Calculate Wave Number (k) ...
Promotional Banner

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

Sound waves travel from air to water. The velocity of the sound waves in air is 332 ms^(-1) and wavelength is 2m. If the wavelength of the sound waves in water is 850 cm, find its velocity in water

Two sound waves of wavelengths 5 m and 6 m formed 30 beats in 3 seconds. The velocity of sound is

Two waves of wavelength 2m and 2.02 m respectively, moving with the same velocity superpose to produce 2 beats/second. The velocity of the waves is

When a sound wave is reflected from a wall the phase difference between the reflected and incident pressure wave is:

A strain of sound waves is propagated along an organ pipe and gets reflected from an open end . If the displacement amplitude of the waves (incident and reflected) are 0.002 cm , the frequency is 1000 Hz and wavelength is 40 cm . Then , the displacement amplitude of vibration at a point at distance 10 cm from the open end , inside the pipe is

NARAYNA-WAVES-Exercise-I (C.W)
  1. Transverse waves are generated in two uniform wires A and B of the sam...

    Text Solution

    |

  2. A string is stretched between fixed points separated by 75.0cm. It is ...

    Text Solution

    |

  3. A sound wave with an amplitude of 3cm starts towards right from origin...

    Text Solution

    |

  4. Sound signal is sent through a composite tube as shown in the figure. ...

    Text Solution

    |

  5. Four simple harmonic vibrations y(1)=8 sin omega t, y(2)= 6 sin (ome...

    Text Solution

    |

  6. A wave pulse on a string has the dimension shown in figure. The w...

    Text Solution

    |

  7. The length of a sonometer wire is 90cm and the stationary wave setup i...

    Text Solution

    |

  8. A sonometer is set on the floor of a lift. When the lift is at rest, t...

    Text Solution

    |

  9. Standing wave produced in a metal rod of length 1m is represented by t...

    Text Solution

    |

  10. An addiditional bridge is kept below a sonometer wire so that it is di...

    Text Solution

    |

  11. A piano wire 0.5m long and mass 5gm is streteched by a tension of 400N...

    Text Solution

    |

  12. An iron load of 2Kg is suspended in the air from the free end of a son...

    Text Solution

    |

  13. The third overtone of a closed pipe is found to be in unison with the ...

    Text Solution

    |

  14. Two closed organ pipes of length 100 cm and 101 cm 16 beats is 20 sec....

    Text Solution

    |

  15. A cylinder resonance tube open at both ends has fundamental frequency ...

    Text Solution

    |

  16. A closed organ pipe is vibrating in first overtone and is in resonance...

    Text Solution

    |

  17. A glass tube of 1.0 m length is filled with water. The water can be dr...

    Text Solution

    |

  18. An open and a closed pipe have same length . The ratio of frequency of...

    Text Solution

    |

  19. A tube of a certain diameter and of length 48cm is open at both ends. ...

    Text Solution

    |

  20. A closed organ pipe has length l. The air in it is vibrating in 3rd ov...

    Text Solution

    |