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Next: Worked example 11.3: Block Up: Oscillatory motion Previous: Worked example 11.1: Piston

Worked example 11.2: Block and spring

Question: A block attached to a spring executes simple harmonic motion in a horizontal plane with an amplitude of $0.25 {\rm m}$. At a point $0.15 {\rm m}$ away from the equilibrium position, the velocity of the block is $0.75 {\rm m/s}$. What is the period of oscillation of the block?

Answer: The equation of simple harmonic motion is

\begin{displaymath}
x = a \cos(\omega t-\phi),
\end{displaymath}

where $x$ is the displacement, and $a$ is the amplitude. We are told that $a=0.25 {\rm m}$. The velocity of the block is obtained by taking the time derivative of the above expression:

\begin{displaymath}
\dot{x} =-a \omega \sin(\omega t -\phi).
\end{displaymath}

We are told that at $t=0$ (say), $x = 0.15 {\rm m}$ and $\dot{x} = 0.75 {\rm m/s}$. Hence,
$\displaystyle 0.15$ $\textstyle =$ $\displaystyle 0.25 \cos(\phi),$  
$\displaystyle 0.75$ $\textstyle =$ $\displaystyle 0.25 \omega \sin(\phi).$  

The first equation gives $\phi = \cos^{-1}(0.15/0.25) = 53.13^\circ$. The second equation yields

\begin{displaymath}
\omega = \frac{0.75}{0.25\times\sin(53.13^\circ)} = 3.75 {\rm rad./s}.
\end{displaymath}

Hence, the period of the motion is

\begin{displaymath}
T = \frac{2 \pi}{\omega} = 1.676 {\rm s}.
\end{displaymath}


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Next: Worked example 11.3: Block Up: Oscillatory motion Previous: Worked example 11.1: Piston
Richard Fitzpatrick 2006-02-02