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frutty [35]
4 years ago
5

A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s

Physics
1 answer:
9966 [12]4 years ago
3 0

Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m

The velocity of the wave in the string is given by :

v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s

So, the velocity of the waves in the string is 72.54 m/s.

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A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
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Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

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= 2.20 × 0.706825 mm

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so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

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3 years ago
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