A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the wavelength (in cm) of the first harmonic?

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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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A string is stretched and fixed at both
ends, 200 cm apart. If the density of
the string is 0.015 g/cm, and its
tension is 600 N, what is the
wavelength (in cm) of the first
harmonic?
600
800
1 000
400
O 200
Transcribed Image Text:A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the wavelength (in cm) of the first harmonic? 600 800 1 000 400 O 200
Expert Solution
Step 1

When a string is stretched and fixed at both ends, it can vibrate to produce a wave. The wave travels along the length of the string and reflects at the fixed ends. This creates standing waves on the string, where certain points on the string do not move while others oscillate with maximum amplitude.

In general, the condition for the nth harmonic of a standing wave on a string is that the string must vibrate with n-1 nodes and n antinodes. The distance between any two adjacent nodes or antinodes is equal to λ/2, where λ is the wavelength of the wave. The wavelength of the nth harmonic is given by:

λ_n = 2L/n

where L is the length of the string and n is the harmonic number.

In this problem, the condition for the wavelength is that it is equal to twice the distance between the fixed ends of the string. This is because for the first harmonic of a standing wave on a string, the wave must have a node (point of no oscillation) at each end of the string, and an antinode (point of maximum oscillation) at the middle of the string. This creates a pattern where the string is divided into two equal segments, each with a length equal to half the wavelength. Therefore, the wavelength of the first harmonic is equal to twice the distance between the fixed ends of the string.

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