Question

A wooden block (m = 0.640 kg) is connected to a spring and undergoes simple harmonic motion with an amplitude of oscillation of 0.055 m. The frequency of the motion is 11.50 Hz. (a) What is the spring constant? N/m (b) What is the maximum speed of the block? m/s (c) What is the speed of the block when it is 0.023 m away from the equilibrium position? m/s

Answer 1

a)

f = frequency = 11.50 Hz

k = spring constant

m = mass = 0.640 kg

A = amplitude = 0.055 m

frequency is given as

f = 1/(2 $\pi$ ) sqrt(k/m)

11.50 = (1/(2 x 3.14)) sqrt(k/0.64)

k = 3338.1 N/m

b)

maximum speed is given as

V_max = A (2 $\pi$ f) = (0.055) (2 x 3.14) (11.50) = 3.97 m/s

c)

x = 0.023 m

using conservation of energy

(0.5) k A² = (0.5) k x² + (0.5) m v²

k A² = k x² + m v²

(3338.1) (0.055)² = (3338.1) (0.023)² + (0.64) v²

v = 3.61 m/s