Q7.5 Finding the amplitude and phase: initial conditions (S)

A particle oscillates in SHM at an angular frequency ω=10rads-1 It is given an initial displacement of 0.5 m and an initial velocity of 2 ms-1. Find the amplitude and the phase constant, and obtain an expression for the displacement as a function of time.
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Hint

You will need Equation 6.5. Think through it before you use it.
 

Solution

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Solution

Since the system is exhibiting SHM its displacement must follow an equation of the general form

x=xmcos(ωt+φ)

Our task is to find out the values that have to be assigned to the amplitude xm and the initial phase φ, on the basis of the information we are given about the displacement and the velocity at time t=0. So we will need the equation for the velocity, which follows by differentiation:

\[ \dot{x} = - x_m\omega \sin(\omega t + \phi) \]

Setting t=0 in the two equations gives:

\[ x(0) = x_m\cos\phi \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} v(0)=\dot{x} = -x_m\omega\sin\phi \]
These are two simultaneous equations in two unknowns (xm and φ). So we can solve them. Dividing one by the other we find:

\[ \frac{v(0)}{x(0)} = -\omega\tan\phi \]

Inserting ω=10rads-1, x=0.5 m and v(0)=2 ms-1 we find

φ=arctan(-0.4)=-0.3805 rad

Now that we have φ, it is straightforward to calculate xm. Using the value of x(0) we find:

xm=x(0)/cosφ=0.54 m
Alternatively, using the value of v(0) we obtain
xm=-v(0)/ωsinφ=0.54 m
which is reassuring.

An explicit equation for the displacement as a function of time now follows by inserting the values for the amplitude and phase into the general solution we began with:

x=(0.54)cos[10t-0.3805]
where x is in metres and t is in seconds.