Question

A 0.45-kg mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass passes through the equilibrium position and the position of the mass at any time is shown in the drawing. Determine the following. (a) amplitude A of the motion m (b) angular frequency rad/s (c) spring constant k N/m (d) speed of the object at t = 2.0 s m/s (e) magnitude of the object's acceleration at t = 2.0 s m/s2

Answer 1

(a) Amplitude A = 0.2 m

(b) Angular Frequency
`\( \omega = 4 \)` rad/s

(c) Spring Constant k = 4 N/m

(d) Speed at
`\( t = 2.0 \) s \( = 0.4 \)` m/s

(e) Acceleration at t = 2.0 s = 0.8 m/s²

Step-by-step explanation:

The amplitude (A) of the motion is the maximum displacement from the equilibrium position. In the given drawing, the amplitude is the distance from the equilibrium position to the maximum displacement, which is 0.2 meters.

The angular frequency
`(\( \omega \))` can be determined from the period of oscillation. Since the period
`(\( T \))` is the time taken for one complete oscillation and is given by
`\( T = (2\pi)/(\omega) \)`, solving for
`\( \omega \)` yields
`\( \omega = 4 \)` rad/s.

The spring constant (k) is related to the angular frequency by
`\( \omega = \sqrt{(k)/(m)} \)`, where m is the mass. Solving for k gives k = 4 N/m.

To find the speed at t = 2.0 s, we can use the equation
`\( v = A\omega\cos(\omega t) \)`, where A is the amplitude and
`\( \omega \)` is the angular frequency. Substituting the given values, we get v = 0.4 m/s.

Finally, the magnitude of acceleration at t = 2.0s is given by
`\( a = A\omega^2\sin(\omega t) \)`. Substituting the known values, we find \( a = 0.8 \) m/s². These calculations provide a comprehensive understanding of the motion of the mass attached to the spring at various points in time.