Home
Class 11
PHYSICS
A rod of nickel of length l is clamped a...

A rod of nickel of length `l` is clamped at its midpoint . The rod is stuck and vibrations are set up in the rod . Find the general expression for the frequency of the longitudinal vibrations of the rod . Young's modulus and density of the rod is `Y and rho` , respectively.

Text Solution

AI Generated Solution

The correct Answer is:
To find the general expression for the frequency of the longitudinal vibrations of a nickel rod clamped at its midpoint, we can follow these steps: ### Step 1: Understand the Setup The rod has a length \( L \) and is clamped at its midpoint. This means that the rod can vibrate in modes where there are nodes at the midpoint. ### Step 2: Determine the Wavelength for the Fundamental Mode For the fundamental mode of vibration in a rod clamped at its midpoint, the length of the rod \( L \) is equal to half the wavelength (\( \lambda_1 \)) of the fundamental frequency. Thus, we can write: \[ L = \frac{\lambda_1}{2} \] This implies: \[ \lambda_1 = 2L \] ### Step 3: Relate Frequency, Wavelength, and Wave Speed The speed of longitudinal waves in a rod is given by: \[ v = \sqrt{\frac{Y}{\rho}} \] where \( Y \) is the Young's modulus and \( \rho \) is the density of the rod. The relationship between speed, frequency, and wavelength is given by: \[ v = f \lambda \] For the fundamental frequency \( f_1 \): \[ v = f_1 \lambda_1 \] ### Step 4: Substitute for Wavelength Substituting \( \lambda_1 = 2L \) into the wave speed equation gives: \[ v = f_1 (2L) \] Thus, we can express the frequency as: \[ f_1 = \frac{v}{2L} \] ### Step 5: Substitute the Speed of Sound in the Rod Now, substituting the expression for \( v \): \[ f_1 = \frac{1}{2L} \sqrt{\frac{Y}{\rho}} \] ### Step 6: Generalize for Higher Modes For the \( n \)-th mode of vibration, the relationship changes. The general expression for the frequency of the \( n \)-th mode can be given by: \[ f_n = \frac{(2n-1)}{2L} \sqrt{\frac{Y}{\rho}} \] where \( n \) is a positive integer (1, 2, 3, ...). ### Final Expression Thus, the general expression for the frequency of the longitudinal vibrations of the rod is: \[ f_n = \frac{(2n-1)}{2L} \sqrt{\frac{Y}{\rho}} \]

To find the general expression for the frequency of the longitudinal vibrations of a nickel rod clamped at its midpoint, we can follow these steps: ### Step 1: Understand the Setup The rod has a length \( L \) and is clamped at its midpoint. This means that the rod can vibrate in modes where there are nodes at the midpoint. ### Step 2: Determine the Wavelength for the Fundamental Mode For the fundamental mode of vibration in a rod clamped at its midpoint, the length of the rod \( L \) is equal to half the wavelength (\( \lambda_1 \)) of the fundamental frequency. Thus, we can write: \[ ...
Promotional Banner

Topper's Solved these Questions

  • SUPERPOSITION AND STANDING WAVES

    CENGAGE PHYSICS|Exercise Single Correct|144 Videos
  • SUPERPOSITION AND STANDING WAVES

    CENGAGE PHYSICS|Exercise Multiple|26 Videos
  • SUPERPOSITION AND STANDING WAVES

    CENGAGE PHYSICS|Exercise Exercise 7.2|32 Videos
  • SOUND WAVES AND DOPPLER EFFECT

    CENGAGE PHYSICS|Exercise Integer|16 Videos
  • THERMODYNAMICS

    CENGAGE PHYSICS|Exercise 24|1 Videos

Similar Questions

Explore conceptually related problems

A steel rod 100 cm long is clamped at its midpoint. The fundamental frequency of longitudinal vibrtions of the rod is 3 kHz. What is the speed of the sound in the rod?

Determine the possible harmonics in the longitudinal vibrations of a rod clampred in the middle.

A steel rod 100 cm long is clamped at its middle. The fundamental frequency of loungitudinal vibrations of the fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 KHz. What is the speed of soind in steel?

A steel rod of length 100 cm is clamped at the middle. The frequency of the fundamental mode for the longitudinal vibrations of the rod is (Speed of sound in steel = 5 km s^(-1) )

A copper rod of length l=50 cm is clamped at its midpoint. Find the number of natural longitudinal oscillations of the rod in the frequency range from 20 to 50 k Hz . What are those frequencies equal to ?

A metallic rod of length 1m has one end free and other end rigidly clamped. Longitudinal stationary waves are set up in the rod in such a way that there are total six antinodes present along the rod. The amplitude of an antinode is 4 xx 10^(-6) m . young's modulus and density of the rod are 6.4 xx 10^(10) N//m^(2) and 4 xx 10^(3) Kg//m^(3) respectively. Consider the free end to be at origin and at t=0 particles at free end are at positive extreme. The magnitude of strain at midpoint of the rod at t = 1 sec is

A metallic rod of length 1m has one end free and other end rigidly clamped. Longitudinal stationary waves are set up in the rod in such a way that there are total six antinodes present along the rod. The amplitude of an antinode is 4 xx 10^(-6) m . young's modulus and density of the rod are 6.4 xx 10^(10) N//m^(2) and 4 xx 10^(3) Kg//m^(3) respectively. Consider the free end to be at origin and at t=0 particles at free end are at positive extreme. The equation describing stress developed in the rod is

A rod AB of mass M and length L is shown in figure. End A of rod is hinged and end B is lying on the ground. Find the Normal reaction by ground on the rod

CENGAGE PHYSICS-SUPERPOSITION AND STANDING WAVES-Subjective
  1. Sound from coherent sources S(1) and S(2) are sent in phase and detect...

    Text Solution

    |

  2. A bat emits ultrasonic sound of frequency 1000 kHz in air . If the sou...

    Text Solution

    |

  3. Figure 7.75 shows a tube structure in which a sound signal is bent fro...

    Text Solution

    |

  4. A source emitting sound of frequency 180 Hz is placed in front of a wa...

    Text Solution

    |

  5. Two coherent narrow slits emitting wavelength lambda in the same phase...

    Text Solution

    |

  6. The following equation represents standing wave set up in a medium , ...

    Text Solution

    |

  7. A wave is given by the equation y = 10 sin 2 pi (100 t - 0.02 x) + ...

    Text Solution

    |

  8. A set of 56 tuning forks is arranged in a sequence of increasing frequ...

    Text Solution

    |

  9. Two tuning forks A and B are sounded together and 8 beats//s are heard...

    Text Solution

    |

  10. A certain fork is found to give 2 beats//s when sounded in conjuction...

    Text Solution

    |

  11. The two parts of a sonometer wire divided by a movable knife edge , di...

    Text Solution

    |

  12. Two tuning forks A and B give 18 beats "in" 2 s. A resonates with one ...

    Text Solution

    |

  13. Six antinodes are observed in the air column when a standing wave form...

    Text Solution

    |

  14. A column of air at 51^(@) C and a tuning fork produce 4 beats per seco...

    Text Solution

    |

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

    Text Solution

    |

  16. A sonometer wire under tension of 128 N vibrates in resonance with a t...

    Text Solution

    |

  17. A rod of nickel of length l is clamped at its midpoint . The rod is st...

    Text Solution

    |

  18. A string is stretched by a block going over a pulley . The string vibr...

    Text Solution

    |

  19. An audio oscillator capable of producing notes of frequencies ranging ...

    Text Solution

    |

  20. A closed orgain pipe of length l(0) is resonating in 5^(th) harmonic m...

    Text Solution

    |