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A stretched string is fixed at both ends which are 160 cm apart. If the density of the string is 0.038 g/cm, and its tension is 600 N, what is the wavelength of the 6th harmonic?

User TabbyCool
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Final answer:

The wavelength of the 6th harmonic of a stretched string fixed at both ends with a length of 160 cm is 53.33 cm.

Step-by-step explanation:

To calculate the wavelength of the 6th harmonic of a stretched string fixed at both ends, we need to use the formula for the wavelength of standing waves on a string. The formula for the nth harmonic on a string of length L is given by:

\[\lambda = \frac{2L}{n}\]

In this case, the length L is 160 cm, and the harmonic number n is 6.

So, the wavelength for the 6th harmonic is:

\[\lambda_6 = \frac{2 \times 160}{6}\]

\[\lambda_6 = \frac{320}{6}\]

\[\lambda_6 = 53.33 \text{ cm}\]

Therefore, the wavelength of the 6th harmonic is 53.33 cm.

User Abdelrhman Arnos
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