Question

and an ideal spring of unknown force constant. The oscillator is found to have a period of 0.157 s and a maximum speed of 2 m/s . You may want to review . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of SHM on an air track, II: Period and frequency.

Answer 1

Answer:

k = 1073.09 N/m

A = 0.05 m

Explanation:

Given:

- Time period T = 0.147 s

- maximum speed V_max = 2 m/s

- mass of the block m = 0.67 kg

Find:

- The spring constant k

- The amplitude of the motion A.

Solution:

- A general simple harmonic motion is modeled by:

                                     x (t) = A*sin(w*t)

- The velocity of the above modeled SHM is:

                                     v = dx / dt

                                     v(t) = A*w*cos(w*t)

- Where A is the amplitude in meters, w is the angular speed rad/s and time t is in seconds.

- We can see that maximum velocity occurs when (cos(w*t)) maximizes i.e it is equal to 1 or -1. Hence,

-                                      V_max = A*w

- Where w is related to mass of the object and spring constant k as follows,

                                       w = sqrt ( k / m )

- The relationship between w angular speed and Time period T is:

                                       w = 2*pi / T

- Equating the above two equations we have,

                                        m*(2*pi / T)^2 = k

- Hence,                           k = 0.67*(2*pi / 0.157)^2

                                        k = 1073.09 N / m

- So, amplitude A is:

                                        A = V_max*sqrt ( m / k )

                                        A = 2*sqrt ( 0.67 / 1073.09 )

                                        A = 0.05 m