# A 100 g ball attached to a spring with spring constant 2.3 N/m oscillates horizontally on a frictionless table. Its velocity is 19 cm/s when x = -5 cm What is the amplitude of oscillation? What is the speed of the ball when x = 3.0 cm? The angular frequency of oscillation is

`omega = sqrt(k/m) =sqrt(2.3/0.1) =4.796 s^-1`

The equation of oscillator motion is

`x(t) = A*sin(omega*t+ phi)`

The speed of the oscillator is

`v(t) =omega*A*cos(omega*t+phi)`

Thus

`(x(t))/(v(t)) = 1/omega*tan(omega*t+ phi)`

`tan(omega*t+phi) = omega*(x(t))/(v(t)) =4.796*(-5)/19 =-1.262`

`omega*t +phi =-51.608 deg`

Amplitude `A` is

`A = x(t)/sin(omega*t+phi)...

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The angular frequency of oscillation is

`omega = sqrt(k/m) =sqrt(2.3/0.1) =4.796 s^-1`

The equation of oscillator motion is

`x(t) = A*sin(omega*t+ phi)`

The speed of the oscillator is

`v(t) =omega*A*cos(omega*t+phi)`

Thus

`(x(t))/(v(t)) = 1/omega*tan(omega*t+ phi)`

`tan(omega*t+phi) = omega*(x(t))/(v(t)) =4.796*(-5)/19 =-1.262`

`omega*t +phi =-51.608 deg`

Amplitude `A` is

`A = x(t)/sin(omega*t+phi) =(-5)/(-0.7838) =6.38 cm`

When `x(t_1) =3` we have

`3 =6.38*sin(omega*t_1 +phi)`

`omega*t_1+phi =28.048 deg`

Thus the speed at time `t_1` is just

`v(t_1) = omega*A*cos(omega*t_1+phi) =4.796*6.38*cos(28.048) =27 (cm)/s`

Answer: the amplitude of oscillator is `A =6.38 cm` and the speed when `x =3 cm` is `v =27 (cm)/s` .