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A 0.50mkg mass at the end of a spring oscillates in simple harmonic motion with an amplitude of 0.15m. If the mass has a maximum speed of 1.25 m/s as it passes the equilibrium position of the spring, then determine the following quantities. a. the spring constant, b. the maximum acceleration, c. the frequency, f, of the oscillation, d. the total energy of the mass spring system.

Respuesta :

Answer:

Explanation:

m = 0.5 kg

A = 0.15 m

vmax = 1.25 m/s

vmax = ω x A

1.25 = ω x 0.15

ω = 8.33 rad/s

(a) Let the spring constant be k

k = ω² m = 8.33 x 8.33 x 0.5 = 34.7 N/m

(b) Maximum acceleration, a max = ω² A = 8.33 x 8.33 x 0.15 = 10.42 m/s^2

(c) Let f be the frequency

ω = 2 π f

8.33 = 2 x 3.14 x f

f = 1.326 Hz

(d) Total energy

E = 1 /2 m x ω² x A² = 0.5 x 0.5 x 8.33 x 8.33 x 0.15 x 0.15 = 0.39 J

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