Question

A stretched string fixed at each end has a mass of 46.0 g and a length of 9.00 m. The tension in the string is 52.0 N.

(a) Determine the positions of the nodes and antinodes for the
third harmonic. (Enter your answers from smallest to largest
distance from one end of the string.)

nodes:

m |

m |

m |

m |

antinodes:

m |

m |

m |

(b) What is the vibration frequency for this harmonic?

Hz

Answer #1

Given that,

Length of the wire = 9.0 m

For the third harmonic frequency, there are 3 loops in the wave

So the length of each loop is

λ/2 = 9/3 => λ = 6

Wave length λ = 6 m

a) Since there are three loops in the third harmonic,

The nodes are occur at the distances 0, λ/2, λ, 3λ/2

i.e. the nodes are occur at **0, 3 m, 6 m, 9
m**

The anti-nodes will occur halfway between each pair of adjacent nodes so that their positions will be at

**1.5 m** [(0+3)/2 = 1.5, **4.5 m**
[(3+6)/2 = 4.5] and **7.5 m** [(6+9)/2 = 7.5]
successively from one end

b) The vibration frequency is F = v/λ

Linear density, μ = m/l = 0.046/9 = 0.0051 kg/m

Given that Tension force T = 52.0 N

Velocity of the wave, v = (T/μ)^(1/2) = 100.98 m/s

F = v/λ = 100.98/6 = 16.83 Hz

Therefore, the vibration frequency is 16.83 Hz

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