Question

A mass 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.

Answer 1

The angular frequency of the oscillation is approximately 9.49 rad/s, and the phase constant φ is approximately 5.6398 radians, assuming the amplitude is equal to the initial displacement.

Step-by-step explanation:

The student has posed a problem related to the simple harmonic motion (SHM) of a mass hanging on a spring. Specifically, they have a mass of 1.5 kg on a spring with a spring constant of 135 N/m and have observed the mass in motion.

Part (a): Angular Frequency (ω):

To find the angular frequency (ω) of the oscillation, we use the formula:

ω = √(k/m)

Substituting in the given values:

ω = √(135 N/m / 1.5 kg)

ω = √(90 s-2) = 9.49 rad/s approximately.

Part (b): Phase Constant (φ):

Given that at t = 0 the mass is at a distance d = 0.45 m below its equilibrium height and moving upward, we can use the initial conditions to find φ. The position is given by y(0) = A cos(φ), and the speed by v(0) = -A ω sin(φ). From the speed equation, sin(φ) = -v(0)/(A ω). Since no amplitude A is given, we assume A = d, which means sin(φ) = -5 m/s / (0.45 m × 9.49 rad/s). From this, we can calculate φ. φ = arcsin(-0.587) approximating, φ lies in the fourth quadrant because the initial speed is positive (upward) and the sine of the angle is negative, φ = 2π - arcsin(0.587) = 2π - 0.6435 rad = 5.6398 rad approximately.

Answer 2

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

(b). The value of Ф is 0.8644 radian.

Step-by-step explanation:

Given that,

Mass m= 1.5 kg

Spring constant = 135 N/m

Distance = 0.45 m

Speed = 5 m/s

`y_(0)=0.45`

The position function is

`y(t)=A\cos(\omega t-\phi)`

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

Using formula of angular frequency

`\omega=\sqrt{(k)/(m)}`

Where, k = spring constant

m = mass

Put the value into the formula

`\omega=\sqrt{(135)/(1.5)}`

`\omega=9.48\ rad/sec`

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

Using given position function

`y(t)=A\cos(\omega t-\phi)`

`y'(t)=v(t)=-A\omega\sin(\omega t-\phi)`

At t = 0,

`y(0)=A\cos(-\phi)`

`0.45=A\cos(-\phi)`...(I)

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

`5=-A\omega\sin(-\phi)`....(II)

Divide equation (II) by equation (I)

`(5)/(0.45)=9.48\tan\phi`

`\phi=\tan^(-1)((5)/(0.45*9.48))`

`\phi=0.8644\ radian`

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

(b). The value of Ф is 0.8644 radian.