Home
Class 12
PHYSICS
A string of length 'l' is fixed at both ...

A string of length `'l'` is fixed at both ends. It is vibrating in its `3^(rd)` overtone with maximum ampltiude `'a'`. The amplitude at a distance `(l)/(3)` from one end is `= sqrt(p)(a)/(2)`. Find `p`.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will analyze the vibration of the string and find the required value of \( p \). ### Step 1: Understand the Overtone The string is vibrating in its 3rd overtone. For a string fixed at both ends, the relationship between the overtone number \( n \) and the wavelength \( \lambda \) is given by: \[ L = n \cdot \frac{\lambda}{2} \] For the 3rd overtone, \( n = 4 \) (since the fundamental mode is the 1st overtone, the 2nd overtone is the 3rd mode, and so on). Thus: \[ L = 4 \cdot \frac{\lambda}{2} \implies \lambda = \frac{L}{2} \] ### Step 2: Write the Expression for Amplitude The amplitude of the wave at a distance \( x \) from one end is given by: \[ A(x) = A \sin(kx) \] where \( k \) is the wave number defined as: \[ k = \frac{2\pi}{\lambda} \] Substituting the value of \( \lambda \): \[ k = \frac{2\pi}{L/2} = \frac{4\pi}{L} \] ### Step 3: Calculate Amplitude at \( x = \frac{L}{3} \) Now, we need to find the amplitude at a distance \( \frac{L}{3} \): \[ A\left(\frac{L}{3}\right) = A \sin\left(k \cdot \frac{L}{3}\right) = A \sin\left(\frac{4\pi}{L} \cdot \frac{L}{3}\right) = A \sin\left(\frac{4\pi}{3}\right) \] ### Step 4: Evaluate \( \sin\left(\frac{4\pi}{3}\right) \) The value of \( \sin\left(\frac{4\pi}{3}\right) \) is: \[ \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} \] Thus, the amplitude becomes: \[ A\left(\frac{L}{3}\right) = A \left(-\frac{\sqrt{3}}{2}\right) \] ### Step 5: Set Up the Equation According to the problem, the amplitude at this point is also given by: \[ A\left(\frac{L}{3}\right) = \frac{\sqrt{p} \cdot A}{2} \] Setting the two expressions for amplitude equal to each other: \[ -\frac{\sqrt{3}}{2} A = \frac{\sqrt{p} \cdot A}{2} \] ### Step 6: Simplify and Solve for \( p \) Dividing both sides by \( A \) (assuming \( A \neq 0 \)): \[ -\frac{\sqrt{3}}{2} = \frac{\sqrt{p}}{2} \] Multiplying both sides by 2: \[ -\sqrt{3} = \sqrt{p} \] Squaring both sides: \[ p = 3 \] ### Final Answer Thus, the value of \( p \) is: \[ \boxed{3} \]

To solve the problem step by step, we will analyze the vibration of the string and find the required value of \( p \). ### Step 1: Understand the Overtone The string is vibrating in its 3rd overtone. For a string fixed at both ends, the relationship between the overtone number \( n \) and the wavelength \( \lambda \) is given by: \[ L = n \cdot \frac{\lambda}{2} \] For the 3rd overtone, \( n = 4 \) (since the fundamental mode is the 1st overtone, the 2nd overtone is the 3rd mode, and so on). Thus: ...
Promotional Banner

Topper's Solved these Questions

  • TRAVELLING WAVES

    RESONANCE|Exercise Exercise- 2 PART III|15 Videos
  • TRAVELLING WAVES

    RESONANCE|Exercise Exercise- 2 PART IV|9 Videos
  • TRAVELLING WAVES

    RESONANCE|Exercise Exercise- 2 PART I|21 Videos
  • TEST SERIES

    RESONANCE|Exercise PHYSICS|127 Videos
  • WAVE ON STRING

    RESONANCE|Exercise Exercise- 3 PART I|19 Videos

Similar Questions

Explore conceptually related problems

A string of length L is fixed at both ends . It is vibrating in its 3rd overtone with maximum amplitude a. The amplitude at a distance L//3 from one end is

An open organ pipe has length l .The air in it vibrating in 3^( rd ) overtone with maximum amplitude A . The amplitude at a distance of (l)/(16) from any open end is.

An open organ pipe has length l .The air in it vibrating in 3^( rd ) overtone with maximum amplitude A . The amplitude at a distance of (l)/(16) from any open end is

An open organ pipe has length l .The air in it vibrating in 3^( rd ) overtone with maximum amplitude A.The amplitude at a distance of (l)/(6) from any open end is.

A string of length L fixed at both ends vibrates in its first overtone. Then the wavelength will be

A closed organ pipe has length L . The air in it is vibrating in thirf overtone with maximum ampulitude a . The amplitude at distance (L)/(7) from closed of the pipe is

A string of length l is fixed at both ends and is vibrating in second harmonic. The amplitude at anti-node is 2 mm. The amplitude of a particle at distance l//8 from the fixed end is

RESONANCE-TRAVELLING WAVES-Exercise- 2 PART II
  1. A certain transvers sinusoidal wave of wavelength 20 cm is moving in t...

    Text Solution

    |

  2. Figure shows a string of linear mass density 1.0 g//cm on which a wave...

    Text Solution

    |

  3. A wire of 9.8 xx 10^(-3) kg per meter mass passes over a fricationless...

    Text Solution

    |

  4. A uniform rope of elngth l and mass m hangs vertically from a rigid su...

    Text Solution

    |

  5. A non-uniform rope of mass M and length L has a variable linear mass d...

    Text Solution

    |

  6. A man generates a symmetrical pulse in a string by moving his hand up ...

    Text Solution

    |

  7. A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported ...

    Text Solution

    |

  8. A string of length 'l' is fixed at both ends. It is vibrating in its 3...

    Text Solution

    |

  9. A string of mass 'm' and length l, fixed at both ends is vibrating in ...

    Text Solution

    |

  10. A travelling wave of amplitude 5 A is partically reflected from a boun...

    Text Solution

    |

  11. A 50 cm long wire of mass 20 g suports a mass of 1.6 kg as shown in f...

    Text Solution

    |

  12. A 1 m long rope, having a mass of 40 g, is fixed at one end and is tie...

    Text Solution

    |

  13. In an experiment of standing waves, a string 90 cm long is attached to...

    Text Solution

    |

  14. Three consecutive resonant frequencies of a string are 90, 150 and 210...

    Text Solution

    |

  15. A steel wire of length 1m, mass 0.1kg and uniform cross-sectional area...

    Text Solution

    |

  16. A wire having a linear density of 0.05 gm//c is stretched between two...

    Text Solution

    |

  17. Figure shows a string stretched by a block going over a pulley. The st...

    Text Solution

    |

  18. Figure shows an aluminium wire of length 60 cm joined to a steel wire ...

    Text Solution

    |

  19. A metallic wire with tension T and at temperature 30^(@)C vibrates wit...

    Text Solution

    |