Question

A body is suspended vertically from an ideal spring of spring constant $2.5 \mathrm{N} / \mathrm{m}$. The spring is initially in its relaxed position. The body is then released and oscillates about its equilibrium position. The motion is described by $$y=(4.0 \mathrm{cm}) \sin [(0.70 \mathrm{rad} / \mathrm{s}) t]$$ What is the maximum kinetic energy of the body?

Short answer

Answer: The maximum kinetic energy of the body is \(0.050\mathrm{J}\).

Step-by-step solution

Find the amplitude and angular frequency

In the equation of motion, the amplitude (A) and angular frequency (ω) can be found directly: $$y=(4.0\mathrm{cm})\sin[(0.70 \mathrm{rad}/\mathrm{s})t]$$. From this equation, we can see that the amplitude is \(A=4.0 \mathrm{cm}=0.040\mathrm{m}\) and the angular frequency is \(\omega=0.70 \mathrm{rad/s}\).

Find the mass of the body

We can find the mass of the body by using the relationship between the spring constant (k), the angular frequency (ω), and the mass (m): $$\omega=\sqrt{\frac{k}{m}}$$. We are given the spring constant \(k=2.5 \mathrm{N/m}\). To find the mass, rearrange the equation and solve for m: $$m=\frac{k}{\omega^2}$$. Now, plug in the values for k and ω: $$m=\frac{2.5\mathrm{N/m}}{(0.70\mathrm{rad/s})^2}$$.

Calculate the mass

Calculate the mass using the values provided: $$m=\frac{2.5\mathrm{N/m}}{(0.70\mathrm{rad/s})^2}=5.10\mathrm{kg}$$

Calculate the maximum kinetic energy

Now that we have the mass, amplitude, and angular frequency, we can find the maximum kinetic energy (K_max) using the equation for the maximum kinetic energy of a simple harmonic oscillator: $$K_\text{max}=\frac{1}{2}m\omega^2A^2$$. Plug in the values for m, ω, and A: $$K_\text{max}=\frac{1}{2}(5.10\mathrm{kg})(0.70\mathrm{rad/s})^2(0.040\mathrm{m})^2$$.

Find the maximum kinetic energy

Calculate the maximum kinetic energy using the values provided: $$K_\text{max}=\frac{1}{2}(5.10\mathrm{kg})(0.70\mathrm{rad/s})^2(0.040\mathrm{m})^2=0.050\mathrm{J}$$. The maximum kinetic energy of the body is \(0.050\mathrm{J}\).