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
