Question

A 0.109-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points ㄨㄧ--0.279 m and X2 = 0.499 m. The period of oscillation is 0.557 s. Find the frequency, the equilibrium position, Xeq, the amplitude, A, the maximum speed, Vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot

Answer 1

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Answer 2

The frequency is the inverse of the period, ?$T$.

$$f=\frac{1}{T}$$

Substitute the given value.

$$f=\frac{1}{0.685\text{ s}}=1.46\text{ Hz}$$

The equilibrium position is the midpoint of the two endpoints. Its coordinate is

$$x_{\text{eq}}=\frac{x_1+x_2}{2}$$

Substitute the data.

$$x_{\text{eq}}=\frac{(-0.253\text{ m})+\left(0.471\text{ m}\right)}{2}=0.109\text{ m}$$

The amplitude is half the distance between the endpoints.

$$A=\frac{|x_1-x_2|}{2}$$

Its value is

$$A=\frac{|(-0.253\text{ m})-\left(0.471\text{ m}\right)|}{2}=0.362\text{ m}$$

The maximum speed is given by

$$v_{\text{max}}=2\pi fA=\frac{\pi |x_1-x_2|}{T}$$

$$v_{\text{max}}=\frac{\pi \times |(-0.253\text{ m})-\left(0.471\text{ m}\right)|}{0.685\text{ s}}=3.32\text{ m/s}$$

The maximum magnitude of acceleration is

$$a_{\text{max}}={\left(2\pi f\right)}^{2}A=2|x_1-x_2|{\left(\frac{\pi }{T}\right)}^2$$

Its value is

$$a_{\text{max}}=2\times |(-0.253\text{ m})-\left(0.471\text{ m}\right)|{\left(\frac{\pi }{0.685\text{ s}}\right)}^2=30.5{\text{ m/s}}^2$$

To find the force constant, start with the formula for the period,

$$T=2\pi \sqrt{\frac{m}{k}}$$

where ?$m$ denotes the mass. Now solve for the force constant.

$$k=m{\left(\frac{2\pi }{T}\right)}^2$$

Insert the given values.

$$k=\left(0.105\text{ kg}\right){\left(\frac{2\pi }{0.685\text{ s}}\right)}^2=8.83\text{ N/m}$$

To find the total mechanical energy, use the fact that it is conserved and evaluate it at any point along the motion. At the equilibrium position, the mechanical energy is completely kinetic, with the particle moving at its maximum speed, ?max$v_{\text{max}}$, which you found previously. The total mechanical energy, therefore, equals the maximum kinetic energy.

$$E_{\text{tot}}=\frac{1}{2}m{v}_{\text{max}}^2=\frac{m}{2}{\left(\frac{\pi |x_1-x_2|}{T}\right)}^2$$

Substitute the given values.

?tot=0.105 kg2(?×|(−0.253 m)−(0.471 m)|0.685 s)2=0.579 J