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Lunna [17]
3 years ago
15

Find the frequency of the 4th harmonic waves on a violin string that is 48.0cm long with a mass of 0.300 grams

Physics
1 answer:
Jlenok [28]3 years ago
4 0

Answer:

The frequency of the 4th harmonic of the string is 481.13 Hz.

Explanation:

When a stretch string fixed at both ends is set into vibration, it produces its lowest sound of possible note called the fundamental frequency.  Under certain conditions on the string, higher frequencies called harmonics or overtones can be produced.

The frequency of the forth harmonic is the third overtone of the string and can be determined by:

          f = \frac{2}{L}\sqrt{\frac{T}{m} }

Given that; L = 48.0 cm = 0.48 m,

                 m = 0.3 g = 0.0003 Kg,

                 T = 4.0 N,

         f = \frac{2}{0.48}\sqrt{\frac{4}{0.0003} }

         f = 4.1667 × 115.4701

           = 481.1252

        f = 481.13 Hz

The frequency of the 4th harmonic of the string is 481.13 Hz.

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\omega_f^2 - \omega_i^2 = 2 \alpha \theta

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Solving the equation for \alpha, we find

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The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

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\omega_i = 11.0 rev/s is the initial angular speed

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t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

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Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

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\omega_f^2 - \omega_i^2 = 2 \alpha \theta

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\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

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Now, the Force exerted on the sled along the ramp is:

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