A 55.0 gblock oscillates in SHM on the end of a spring with k = 1500 N/maccording to $\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{m}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{\omega }\mathit{t}\mathbf{+}\mathit{\varphi }\mathbf{)}$. How long does the block take to move from positionto $\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{800}{\mathit{x}}_{\mathbf{m}}$(a) position $\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{600}{\mathit{x}}_{\mathbf{m}}$and (b) position$\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{800}{\mathit{x}}_{\mathbf{m}}$?

• Mass of block,m = 55.0 g or 0.055 kg.
• Spring constant, k = 1500 N/m.
• The equation of displacement,$x=x_m\cos (\omega t+\varphi )$
• Block moves from position, $+0.800x_m$.

Motion is simple harmonic so, using displacement x(t)and calculatingangular frequency,we can find thetime taken bytheblock to move from position$\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{800}{\mathit{x}}_{\mathbf{m}}$toposition$\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{600}{\mathit{x}}_{\mathbf{m}}$and position$\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{800}{\mathit{x}}_{\mathbf{m}}$.

Formula:

The angular frequency of the wave,$\omega =\sqrt{\frac{k}{m}}$ (i)

The displacement equation of the wave, $x=x_m\cos \left(\omega t+\varphi \right)$ (ii)

Using equation (i) and the given values, we get the angular frequency of the oscillations as:

$$\begin{gathered} \omega =\sqrt{\frac{1500}{0.055}} \\ =165.1\mathrm{rad}/\mathrm{s} \end{gathered}$$

Let, the motion from $X_1=+0.800x_m$at time ${\mathrm{t}}_1$to = at time${\mathrm{t}}_2$.

Now, using equation (ii), we can write the first displacement equation as:

$$\begin{gathered} x_1=x_m\cos (\omega t_1+\varphi ) \\ +0.800x_m=x_m\cos (\omega t_1+\varphi ) \\ +0.800=\cos (\omega t_1+\varphi ) \\ (\omega t_1+\varphi )={\cos }^{-1}(0.800) \\ (\omega t_1+\varphi )=0.6435\mathrm{rad} \end{gathered}$$

Similarly, using equation (ii), we can write the second displacement equation as:

localid="1657268928304" $$\begin{gathered} x_2=x_m\cos (\omega t_2+\varphi ) \\ +0.600x_m=x_m\cos (\omega t_2+\varphi ) \\ +0.600=\cos (\omega t_2+\varphi ) \\ (\omega t_2+\varphi )={\cos }^{-1}(0.800) \\ (\omega t_2+\varphi )=0.9272\mathrm{rad} \end{gathered}$$

Subtracting the first equation from second, we get,

localid="1657269058209" $$\begin{gathered} (\omega t_2+\varphi )-(\omega t_2+\varphi )=0.9272-0.6435 \\ \omega (t_2-t_1)=0.9272-0.6435 \\ \omega (t_2-t_1)=0.2837 \\ (t_2-t_1)=\frac{0.2837}{\omega } \\ (t_2-t_1)=\frac{0.2837}{165.1} \\ (t_2-t_1)=1.72\times {10}^{-3}\mathrm{s} \end{gathered}$$

Hence, the required time is $1.72\times {10}^{-3}\mathrm{s}$.

Let, the motion from $x_1=+0.800x_m$at time $t_1$to $x_3=+0.800x_m$at time$t_3$

Using equation (ii), we can write the displacement equation as:

$$\begin{gathered} x_3=x_m\cos (\omega t_3+\varphi ) \\ -0.800x_m=x_m\cos (\omega t_3+\varphi ) \\ -0.800=\cos (\omega t_2+\varphi ) \\ (\omega t_3+\varphi )={\cos }^{-1}(-0.800) \\ (\omega t_3+\varphi )=2.4981\mathrm{rad} \end{gathered}$$

Subtracting this equation from second equation, we get,

$$\begin{gathered} (\omega t_3+\varphi )-(\omega t_1+\varphi )=2.4981-0.6435 \\ \omega (t_3-t_1)=2.4981-0.6435 \\ \omega (t_3-t_1)=1.8546 \\ (t_3-t_1)=\frac{1.8546}{\omega } \\ (t_3-t_1)=\frac{1.8546}{165.1} \\ (t_3-t_1)=11.2\times {10}^{-3}\mathrm{s} \end{gathered}$$

Hence, the required value of time is $11.2\times {10}^{-3}\mathrm{s}$.