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A particle is moving in simple harmonic motion with an amplitude of 3.0 cm and a maximum velocity of 15.0 cm/s. Assume the initial phase is 0. (Include the sign of the value in your answer.) (a) At what position is its velocity 4.4 cm/s? cm (b) What is its velocity when its position is 1.5 cm

Respuesta :

Answer:

a)x=2.86 cm

b)V= 12.99 cm/s

Explanation:

Given that

Amplitude A= 3 cm

Maximum velocity V(max) = 15 cm/s

We know that the  speed in simple harmonic motion given as

[tex]V=\omega\sqrt{A^2-x^2}[/tex]

x=Displacement from mean position

ω =Angular speed

V(max) = ωA

Now by putting the values

15 =  ω x 3

ω = 5 rad/s

a)

V= 4.4 cm/s

[tex]V=\omega\sqrt{A^2-x^2}[/tex]

[tex]4.4=5\times \sqrt{3^2-x^2}[/tex]

x=2.86 cm

b)

x= 1.5 cm

[tex]V=\omega\sqrt{A^2-x^2}[/tex]

[tex]V=5\times \sqrt{3^2-1.5^2}[/tex]

V= 12.99 cm/s