Question

(a) A novelty clock has a 0.0100-kg-mass object bouncing on a spring that has a force constant of 1.25 N/ m. What is the maximum velocity of the object if the object bounces 3.00 cm above and below its equilibrium position

Answer 1

Answer:

0.3354 m/s

Explanation:

Using the general equation of a wave,

x(t) = Acos(ωt+Φ)....................... Equation 1

Where x(t) = distance of the wave at an instantaneous time t, A = maximum displacement of the wave, ω = angular velocity, t = time, Φ = phase angle.

Note: v(t) = x(t)/t, and this is the differentiation of x(t)

V(t) = -Aω sin(ωt+Φ)............. Equation 2

From the equation above,

V(max) = Aω............... Equation 3

But,

ω = √(k/m)................ Equation 4

Where k = force constant of the spring, m = mass of the object.

Given: k = 1.25 N/m, m = 0.01 kg

Substitute these values into equation 4.

ω = √(1.25/0.01)

ω = √125

ω = 11.18 rad/s

Also given: A = 3.00 cm = 0.03 m

Substitute into equation 3

V(max) = 11.18(0.03)

V(max) = 0.3354 m/s