Question

A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.8 cm.
(a) Calculate the maximum value of its speed.

(b) Calculate the maximum value of its acceleration.

(c) Calculate the value of its speed when the object is 6.80 cm from the equilibrium position. cm/s

(d) Calculate the value of its acceleration when the object is 6.80 cm from the equilibrium position. cm/s2

e) Calculate the time interval required for the object to move from x = 0 to x = 2.80 cm.

Answer 1

Answer:

Explanation:

mass, m = 0.520 kg

K = 8 N/m

Amplitude, A = 10.8 cm

(a)

Angular speed, $\omega =\sqrt{\frac{K}{m}}$

$\omega =\sqrt{\frac{8}{0.520}}=3.92$

maximum speed = ωA = 3.92 x 10.8 = 42.34 cm/s

(b) maximum acceleration, a = ω²A = 3.92 x 3.92 x 10.8 = 166 cm/s^2

(c) Speed of the particle is given by

$v=\omega \sqrt{A^{2}-y^{2}}$

$v=3.92 \sqrt{10.8^{2}-6.80^{2}}$

v = 33 cm/s

(d) acceleration

a = ω²y = 3.92 x 3.92 x 6.80 = 104.5 cm/s^2

(e) x = A Sinωt

2.80 = 10.8 Sin 3.92 t

Sin 3.92 t = 0.259

3.92 t = 0.262

t = 0.0668 second