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3 years ago

Select the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz.

1 answer:

6 0

It is defined that a string that is in its two ends, its fundamental frequency and its subsequent frequencies are the product of the whole depending on the number of the harmonic. In other words, harmonics (higher) define the subsequent frequencies under the functions 2f, 3f, 4f, 5f, etc.

Therefore we have that the higher harmonics would be:

1 x 80 Hz = 80 Hz (1st harmonic and Fundalmental Frequency)

2 x 80Hz = 160Hz (2nd harmonic)

3 x 80Hz = 240Hz (3rd harmonic)

4 x 80Hz = 320Hz (4th harmonic)

5 x 80Hz = 400Hz (5th harmonic)

Hence, the frequencies 160Hz and 240Hz are the two higher harmonics of string with a fundamental frequency of 80Hz.

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Explanation:

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$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

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Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

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