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
