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 antinode is 2.20 mm and the speed of propagation of transverse waves on the string is 260 m/s. The string extends along the x-axis, with one of the fixed ends at x= 0, so that there is a node at x =0. The smallest value of x where there is an antinode is x= 0.150m.

Required:
a. What is the maximum transverse speed of a point on the string at an antinode?
b. What is the maximum transverse speed of a point on the string at x = 0.075 m?

Question

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Respuesta :

nuhulawal20 nuhulawal20 · Answer 1 · 2021-02-12T02:45:42+01:00

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 [tex]V_{max1}[/tex] = A × ω

[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s

[tex]V_{max1}[/tex] = 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 )

= 2.20 sin( 0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

[tex]V_{max2}[/tex] = A' × ω

[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69

[tex]V_{max2}[/tex] = 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