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Dimas [21]
3 years ago
11

What are the first three harmonics of a note produced on a 32.0 cm long violin string if waves on this string have a speed of 28

0.0 m/s?
___ Hz
___Hz
___Hz
Physics
1 answer:
vodka [1.7K]3 years ago
6 0

Answer:

The first harmonics are the fundamental, the first overtone, and the second overtone,

For a string fixed at both ends you have:

N-A-N            F/2

N-A-N-A-N     F

N-A-N-A-N-A-N    3 F/2        each harmonic increases by F/2

Where N are the anti-nodes and A the antinodes

N-A-N-A-N    represents 1 wavelength for string fixed at both ends

So 1 wavelength is 64 cm

280 m/s = F * .64 m       and F = 437.5 Hz       (second harmonic)

First harmonic = F/2 = 218.75 / sec

Third harmonic = 3 F / 2 = 656.25 /sec

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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
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Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

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So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

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Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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