Question

A particle with mass oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of and a duration of for cycles of oscillation. Find the frequency, the speed at the equilibrium position, the spring constant, the potential energy at an endpoint, the potential energy when the particle is located of the amplitude away from the equiliibrium position, and the kinetic energy, and the speed, at the same position.

find:

F=

vmax=

k=

U max=

U=

K=

v=

Answer 1

Let be: Amplitude: A Period: T Mass: m The frequency is giving by: f= The speed at the equilibrium point is the maximum speed: A21 Vmax = Aw = I Remember that: w = 1 vm Then, the spring constant is: k = mw2 The potential energy at end point is giving by: U = 3k A2 = }(mw?)A? The potential energy at a amplitude away: U = Umax The kinetic energy at this position: K = 0 v=0

Answer 2

Frequency is the number of cycles per unit time, so

$$f=\frac{n}{t}$$

where ?$n$ denotes the number of cycles counted and ?$t$ is their duration. Insert the data.

$$f=\frac{63\text{ cycles}}{127\text{ s}}=0.496\text{ Hz}$$

The speed at the equilibrium position, which is the maximum speed of the motion, is given by

$$v_{\text{max}}=2\pi fA=2\pi \frac{n}{t}A$$

in terms of given quantities, where ?$A$ denotes the amplitude. Substitute the given values.

$$v_{\text{max}}=2\pi \times \frac{\left(63\text{ cycles}\right)}{\left(127\text{ s}\right)}\left(0.929\text{ m}\right)=2.90\text{ m}/\text{s}$$

For the spring constant, use the frequency formula,

$$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$$

with ?$m$ denoting the mass, and solve for ?$k$ in terms of given quantities.

$$k=m{\left(2\pi f\right)}^2=m{\left(2\pi \frac{n}{t}\right)}^2$$

Enter the numbers.

$$k=\left(1.89\text{ kg}\right){(2\pi \times \frac{63\text{ cycles}}{127\text{ s}})}^2=18.4\text{ N}/\text{m}$$

The potential energy at an endpoint, which is the maximum potential energy of the oscillator, is given by

$$U_{\text{max}}=\frac{1}{2}k{A}^2=2m{\left(\frac{\pi nA}{t}\right)}^2$$

Substitute the values of the quantities.

$$U_{\text{max}}=2\times \left(1.89\text{ kg}\right){\left(\frac{\pi \times \left(63\text{ cycles}\right)\left(0.929\text{ m}\right)}{127\text{ s}}\right)}^2=7.92\text{ J}$$

The potential energy at the given location is

$$U=\frac{1}{2}k{\left(\varphi A\right)}^2=2m{\left(\frac{\pi n\varphi A}{t}\right)}^2$$

where ?$\varphi$ denotes the fraction of the amplitude (converted from the given percentage). Insert the numbers.

$$U=2\times \left(1.89\text{ kg}\right){\left(\frac{\pi \times \left(63\text{ cycles}\right)\times 0.495\times \left(0.929\text{ m}\right)}{127\text{ s}}\right)}^2=1.94\text{ J}$$

The kinetic energy at the given location equals the difference between the total mechanical energy and the potential energy there since the total mechanical energy is the sum of the kinetic and potential energies. The total mechanical energy is conserved, so it can be calculated at any location. At an endpoint, the kinetic energy is zero and the total mechanical energy equals the potential energy there, ?max,$U_{\text{max}}\text{,}$ which was calculated previously.

$$\begin{array}{rl}K& =U_{\text{max}}-U\\ & =\frac{1}{2}k{A}^2-\frac{1}{2}k{\left(\varphi A\right)}^2\\ & =\frac{1}{2}k{A}^2(1-{\varphi }^2)\\ & =2m{\left(\frac{\pi nA}{t}\right)}^2(1-{\varphi }^2)\end{array}$$

Substitute the data.

$$K=2\times \left(1.89\text{ kg}\right){\left(\frac{\pi \times \left(63\text{ cycles}\right)\left(0.929\text{ m}\right)}{127\text{ s}}\right)}^2(1-{0.495}^2)=5.98\text{ J}$$

For the speed at the given location, use the kinetic energy formula,

$$K=\frac{1}{2}m{v}^2$$

and solve for the speed in terms of given quantities.

$$v=\sqrt{\frac{2K}{m}}=\frac{2\pi nA}{t}\sqrt{1-{\varphi }^2}$$

Substitute the given values.

?=2?×(63 cycles)127 s1−0.4952⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√=2.52 m/s