91影视

• The frequency of oscillation is, $\mathrm{f}=0.25\mathrm{Hz}$.
• At the time $t=0$, velocity is, $v=0\mathrm{m}/\mathrm{s}$.
• At $\mathrm{t}=0$, the displacement is, $x=0.37\mathrm{cm}\mathrm{or}0.0037\mathrm{m}$.

As per Hooke鈥檚 law, a particle with mass m moves under the influence of restoring force, $\mathit{F}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}$undergoes a simple harmonic motion. Here, Fis restoring force, kis force constant and x is the displacement from the mean position.

In simple harmonic motion, displacement of the particle is given by the equation,

$\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{m}}\mathbf{cos}\mathbf{(}\mathit{蝇}\mathit{t}\mathbf{+}\mathbf{蠁}\mathbf{)}$

Using the expression for a simple harmonic motion for displacement, velocity, and formulae for$\mathit{T}\mathbf{,}\mathit{蝇}\mathbf{,}{\mathit{V}}_{\mathbf{max}}\mathbf{,}{\mathit{a}}_{\mathbf{max}}$ we can findtherespective values.

Formula:

The angular frequency of oscillation,$蝇=2\mathrm{蟺}f$ (i)

The period of oscillation, $T=\frac{1}{f}$ (ii)

The maximum speed of a body, $v_{max}=蝇x_m$ (iii)

The magnitude of maximum acceleration of a body, $\left|a_{max}\right|={蝇}^2{x}_m$ (iv)

The displacement equation at zero phase, $x\left(t\right)=x_m\cos \left(蝇t\right)$ (v)

The velocity equation at zero phase, $v\left(t\right)=-蝇x_m\sin \left(蝇t\right)$ (vi)

Here, $f$is frequency, $x_m$is maximum displacement or amplitude, $t$ is time.

Using equation (ii), we can get the period of oscillations as:

$$\begin{gathered} T=\frac{1}{0.25\mathrm{Hz}} \\ =4\mathrm{s} \end{gathered}$$

Hence, the value of period is $4\mathrm{s}$.

Using equation (i), we can get the angular frequency of oscillation as:

$$\begin{gathered} 蝇=2脳3.14脳0.25\mathrm{Hz} \\ =1.57\mathrm{rad}/\sec \end{gathered}$$

Hence, the value of angular frequency is role="math" localid="1657257255516" $1.57\mathrm{rad}/\sec$.

As at $x=0.0037\mathrm{m},v=0$and at extreme positions $v=0$

role="math" localid="1657257658285" $$\begin{gathered} \mathrm{Hence},\mathrm{x}={\mathrm{x}}_m \\ =3.7脳{10}^{-3}\mathrm{m} \\ =0.0037\mathrm{m} \end{gathered}$$

Hence, the value of amplitude is $0.0037\mathrm{m}$.

By substituting the given and derived values in equation (v), we get the displacement equation as:

$x\left(t\right)=(0.37\mathrm{cm})\cos (1.57\mathrm{t})$

Hence, the displacement equation is $(0.37\mathrm{cm})\cos (1.57\mathrm{t})$.

By substituting the given and derived values in equation (vi), we get the velocity equation as:

$$\begin{gathered} v\left(t\right)=-1.57\mathrm{rad}/\mathrm{s}脳0.37\mathrm{cm}\sin (1.57t) \\ =-(0.58\mathrm{cm}/\mathrm{s})\sin (1.57t) \end{gathered}$$

Hence, the velocity equation is$-(0.58cm/s)\sin (1.57t)$.

Using equation (iii), we can get the maximum speed of the particle as:

$$\begin{gathered} {\mathrm{v}}_{max}=1.57\mathrm{rad}/\mathrm{s}脳0.37\mathrm{cm} \\ =0.58\mathrm{cm}/\mathrm{s} \end{gathered}$$

Hence, the value of maximum speed is $0.58\mathrm{cm}/\mathrm{s}$.

Using equation (iv), we can get the magnitude of the maximum acceleration as:

$$\begin{gathered} \left|{\mathrm{a}}_{\max }\right|={(1.57\mathrm{rad}/\mathrm{s})}^2脳0.37\mathrm{cm} \\ =0.91\mathrm{cm}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of maximum acceleration is $0.91\mathrm{cm}/{\mathrm{s}}^2$.

Substituting the given values for in equation (v), we get the displacement as:

$$\begin{gathered} \mathrm{x}\left(3\right)=(0.37\mathrm{cm})\cos (1.57\mathrm{rad}/\mathrm{s}脳3\mathrm{s}) \\ =(0.37\mathrm{cm})\cos (4.71\mathrm{rad}) \\ 鈮�0 \end{gathered}$$

Hence, the displacement value is 0 m.

Substituting the given values for in equation (vi), we get the velocity as:

$$\begin{gathered} \mathrm{v}(3s)=-(1.57\mathrm{rad}/\mathrm{s})脳(0.37\mathrm{cm})sin(1.57\mathrm{rad}/\mathrm{s}脳3\mathrm{s}) \\ =(-0.5809\mathrm{cm}/\mathrm{s})脳sin(4.71\mathrm{rad}) \\ =(-0.5809\mathrm{cm}/\mathrm{s})脳(-0.999) \\ =0.58\mathrm{cm}/\mathrm{s} \end{gathered}$$

Hence, the displacement value is 0.58 cm/s.