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.

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