A cart of mass 1.4 kg is attached to a horizontal spring with stiffness 44.4545 N/m and

is able to move horizontally without friction.
The cart is then acted on by a sinusoidal driving force with ω = 4.9 Hz and an amplitude of 1.74532 N.
What is the amplitude of the cart’s horizontal motion under the effects of the driving force?

Answer in units of m

To find the amplitude of the cart's horizontal motion under the effects of the driving force, we need to use the concept of resonance.

Resonance occurs when the frequency of the driving force matches the natural frequency of the system. In this case, the natural frequency is determined by the stiffness of the spring and the mass of the cart.

The natural frequency (ω_n) can be calculated using the formula:

ω_n = sqrt(k / m)

where k is the stiffness of the spring and m is the mass of the cart.

Plugging in the given values:

k = 44.4545 N/m
m = 1.4 kg

ω_n = sqrt(44.4545 / 1.4)
= sqrt(31.7532)
≈ 5.63 Hz

Since the driving force has a frequency (ω) of 4.9 Hz, which is slightly less than the natural frequency, the amplitude of the cart's motion will be less than the amplitude of the driving force.

To find the amplitude of the cart's motion, we use the equation:

A_cart = A_driving_force / sqrt((ω_n^2 - ω^2)^2 + (2ξω)^2)

where A_cart is the amplitude of the cart's motion, A_driving_force is the amplitude of the driving force, ω_n is the natural frequency, ω is the frequency of the driving force, and ξ is the damping factor (assuming no damping in this case).

Plugging in the given values:

A_driving_force = 1.74532 N
ω_n = 5.63 Hz
ω = 4.9 Hz

A_cart = 1.74532 / sqrt((5.63^2 - 4.9^2)^2 + (2 * 0 * 4.9)^2)
= 1.74532 / sqrt((31.7529 - 24.01)^2)
= 1.74532 / sqrt(7.7429^2)
= 1.74532 / 7.7429
≈ 0.2253 m

Therefore, the amplitude of the cart's horizontal motion under the effects of the driving force is approximately 0.2253 m.