Amass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. the spring has spring constant k = 135 n/m and negligible mass. the mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y(t) = a cos(ωt – φ), with the positive y-axis pointing upward. at time t = 0 the mass is observed to be at a distance d = 0.45 m below its equilibrium height with an upward speed of v0 = 5 m/s.
(a) find the angular frequency of the oscillation in radians per second. 20%
(b) find the value of, in radians.

(a). The angular frequency is 9.48 rad/s.

(b). The value of Ф is 0.8644 radian.

Explanation:

Given that,

Mass m= 1.5 kg

Spring constant = 135 N/m

Distance = 0.45 m

Speed = 5 m/s

The position function is

(a). We need to calculate the angular frequency of the oscillation

Using formula of angular frequency

Where, k = spring constant

m = mass

Put the value into the formula

(b). We need to calculate the value of φ

Using given position function

At t = 0,

...(I)

At t = 0, v(0) =5 m/s

....(II)

Divide equation (II) by equation (I)

Hence, (a). The angular frequency is 9.48 rad/s.

(b). The value of Ф is 0.8644 radian.

an energy transformation flow diagram is shown

where is it ?