Question

A \(170-g\) object on a spring oscillates left to right on a frictionless surface with a frequency of \(3.00 \mathrm{Hz}\) and an amplitude of $12.0 \mathrm{cm} .$ (a) What is the spring constant? (b) If the object starts at \(x=12.0 \mathrm{cm}\) at \(t=0\) and the equilibrium point is at \(x=0,\) what equation describes its position as a function of time?

Short answer

Answer: The spring constant (k) is 53.94 N/m. The equation that describes the object's position as a function of time is x(t) = 0.12*cos(18.85*t).

Step-by-step solution

Convert the mass and amplitude to SI units

As we are going to deal with SI units, we need to convert the mass and amplitude to the respective SI units. Mass: 170 g = 0.17 kg Amplitude: 12.0 cm = 0.12 m

Calculate the angular frequency (ω)

The problem provides us with the frequency (f) in Hz, and we can calculate the angular frequency using the formula ω = 2πf. ω = 2π * 3.00 Hz = 18.85 rad/s

Calculate the spring constant (k)

Now, we can use the formula ω = sqrt(k/m) to find the spring constant (k). Rearranging the formula to solve for k: k = ω²*m k = (18.85 rad/s)² * 0.17 kg = 53.94 N/m So, the spring constant is 53.94 N/m.

Determine the phase constant (φ)

Since the object starts at the maximum amplitude position (x = 12.0 cm = 0.12 m) and moves left at t = 0, the phase constant (φ) should be zero, as the cosine function is at its peak at 0. φ = 0

Write the equation for the position as a function of time

Now that we have all the components, we can write the equation for the position of the object as a function of time: x(t) = A*cos(ω*t + φ) x(t) = 0.12*cos(18.85*t) So, the equation that describes the object's position as a function of time is x(t) = 0.12*cos(18.85*t).