Question

The displacement of an oscillating object as a function of time is shown in Fig. E14.4. What are (a) frequency; (b) the amplitude; (c) the period; (d) the angular frequency of this motion?

Short answer

(a) Frequency is_{$6.25x{10}^3\mathrm{Hz}$}.

(b) The amplitude is_$10cm$.

(c) The period is_$16s$.

(d) The angular frequency of this motion is $0.39rad/s$.

Step-by-step solution

Definition of frequency amplitude, period, and angular frequency.

Frequency is defined as the number of oscillations done by an object per second.

The amplitude of motion can be defined as the maximum distance that an object travels before coming back to its original position.

The period can be defined as the time taken required to complete one cycle or oscillation.

Angular frequency is a quantity that measures how quickly an object oscillates with respect to time.

(a) Calculate frequency.

The given figure shows one complete cycle for the object. Thus, the period for one complete cycle from the graph is

T = 16 s

The relation between frequency and time period is given by,

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

Substitute value of in above expression to get frequency

$$\begin{gathered} f=\frac{1}{16s} \\ =0.0625Hz \\ =62.5\times {10}^{-3}Hz \end{gathered}$$

Therefore, the frequency is $62.5\times {10}^{-3}Hz$.

(b) Calculate the amplitude

(b) It is given that the object travels from - 10 cm to +10 cm . Thus, its amplitude is the maximum distance its travels from the equilibrium position.

Therefore, the amplitude is 10 cm .

(c) Determine the period.

(c) In part (a), the time period was calculated as

T = 16 s

Therefore, the time period is 16 s .

(d) Find angular frequency of the motion

The relation between angular frequency and time period is given by,

$\omega =\frac{2\mathrm{\pi }}{\mathrm{T}}$

Substitute the values of time period to get angular frequency

$$\begin{gathered} \omega =\frac{2\mathrm{\pi }}{16s} \\ =0.39rad/s \end{gathered}$$

Therefore, the angular frequency is0.39 rad/s .