Question

A ball whose mass is$\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}\mathit{k}\mathit{g}$ suspended from a spring whose stiffness is $\mathbf{4}\mathbf{}\mathit{N}\mathbf{/}\mathit{m}$. The ball oscillates up and down with an amplitude of $\mathbf{14}\mathbf{}\mathit{c}\mathit{m}$. (a) What is the angular frequencyrole="math" localid="1657731610160" $\mathit{\omega }$? (b) What is the frequency? (c) What is the period? (d) Suppose this apparatus were taken to the moon, where the strength of the gravitational field is onlyrole="math" localid="1657731589019" $\mathbf{1}\mathbf{/}\mathbf{6}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathit{e}\mathit{a}\mathit{r}\mathit{t}\mathit{h}$ . What would be the period of the Moon? (Consider carefully how the period depends on properties of the system, look at the equation.)

Short answer

1. The angular frequency is $2.86rad/s$

2. The frequency is $0.45Hertz$

3. The period is $2.2s$

4. Hence, the time period of Moon is $5.39s$

Step-by-step solution

Identification of the given data

• The mass of a ball is$1.4kg$.

• The stiffness is$4N/m$

• The amplitude is $14cm$

• The strength of the gravitational field is only $1/6ofthatearth$

Concept of stiffness, amplitude

Stiffnessis defined asthe force required producing a unit deformation.

Amplitude is defined as the maximum displacement obtained by the object.

(a) Determination of the angular frequency

Angular Frequency is defined as the angular displacement of any object of the wave in unit time.

Angular Frequency,

$\mathit{\omega }\mathbf{=}\sqrt{\frac{{\mathbf{k}}_{\mathbf{s}}}{\mathbf{m}}\mathbf{}\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

where,

$$\begin{gathered} k_s=Thestiffness \\ m=Themass \end{gathered}$$

Substituting the values in above expression,

$$\begin{gathered} \omega =\sqrt{\frac{4}{1.4}} \\ \omega =2.86rad/s \end{gathered}$$

Hence, the angular frequency is$\mathbf{2}\mathbf{.}\mathbf{86}\mathbf{}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}\mathit{s}$

(b) Determination of the frequency

Frequency is defined as the number of cycles required to complete a full wave per unit time.

Frequency,

$$\begin{gathered} \mathit{f}\mathbf{=}\frac{\mathbf{\omega }}{\mathbf{2}\mathbf{\pi }} \\ =\frac{2.86}{2\pi } \\ =0.45Hertz \end{gathered}$$

Hence, the frequency is $\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{}\mathit{H}\mathit{e}\mathit{r}\mathit{t}\mathit{z}$.

(c) Determination of the period

Time Period is defined as the time required completing a full wave to cross a point.

Period,

$\mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\omega }}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Substituting the value in above expression,

$$\begin{gathered} T=\frac{2\pi }{2.86} \\ T=2.2s \end{gathered}$$

Hence, the period is $\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{}\mathit{s}$.

(d) Determination of the period of the Moon

Time period of the Moon,

$$\begin{gathered} T_{moon}\approx \sqrt{6.}{T}_{earth} \\ Where,T_{earth}=2.2s \\ {T}_{moon}\approx \sqrt{6}\times 2.2s \\ \approx 5.39s \end{gathered}$$

Hence, the time period of Moon is$\mathbf{5}\mathbf{.}\mathbf{39}\mathbf{}\mathit{s}$