91Ó°ÊÓ

• The frequency of the oscillation,$f=1000\mathrm{Hz}$.
• The amplitude of the oscillation, $x_m=0.40\mathrm{mm}$or$0.40×{10}^{-3}\mathrm{m}$ .

Using the expression of SHM for maximum velocity and maximum acceleration we can find maximum velocity and acceleration. Using the expression for velocity and acceleration at a given displacement we can find the value of displacement and acceleration.

Formula:

The maximum speed of the oscillation,$V_{max}=x_m\mathrm{Ó\neg }$ (i)

The magnitude of maximum acceleration of the oscillation,$\left|a_{max}\right|={\mathrm{Ó\neg }}^2{x}_m$ (ii)

The angular frequency of the oscillation, $\mathrm{Ó\neg }=2\mathrm{Ï„}\mathrm{Ï„}f$ (iii)

The value of angular frequency using equation (iii) can be given as:

$\mathrm{Ó\neg }=2×3.14×1000\mathrm{Hz}$

$=6.28×{10}^3\mathrm{rad}/\sec$

Now, the value of the magnitude of maximum acceleration using equation (ii) is given as:

$$\begin{gathered} \left|a_{max}\right|={\left(6.28×{10}^3\mathrm{rad}/\mathrm{s}\right)}^2×\left(0.40×{10}^{-3}\mathrm{m}\right) \\ =15.77×{10}^3\mathrm{m}/{\mathrm{s}}^2 \\ =1.6×{10}^4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value maximum acceleration is $1.6×{10}^4\mathrm{m}/{\mathrm{s}}^2$.

Using equation (i), the value of maximum speed of the oscillations is given as:

$$\begin{gathered} V_{max}=\left(6.28×{10}^3\mathrm{rad}/\mathrm{s}\right)×\left(0.40×{10}^{-3}\mathrm{m}\right) \\ =2.512\mathrm{m}/\mathrm{s} \\ ≈2.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the required value of speed is 2.5 m/s .

Using equation (ii), the required acceleration at the position is given as:

$$\begin{gathered} \left|a_{0.20}\right|=\left(6.28×{10}^3\right)×0.20×{10}^{-3} \\ =7.886×{10}^3\mathrm{m}/{\mathrm{s}}^2 \\ ≈7.9×{10}^3\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of acceleration is $7.9×{10}^3\mathrm{m}/{\mathrm{s}}^2$.

The required value of the speed at the position 0.2 m is given as:

$$\begin{gathered} V_{0.20}=\mathrm{Ó\neg }{x}_m\sqrt{1-{\left(\frac{x}{x_m}\right)}^2} \\ =\left(6.28×{10}^3\right)×0.40×{10}^{-3}\sqrt{1-{\left(\frac{0.20}{0.40}\right)}^2} \\ =2.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of speed is 2.2 m/s .