Question 14.5: A particle is moving with SHM of period π. Initially it is 1...

The information given is shown in the diagram below.

The initial speed is positive, so an appropriate equation is

x = a sin(ωt + ε),

and you need to find the values of a, ω, and ε.

Finding ω
Since the period of the motion is π,

\frac{2π}{ω} = π    ⇒    ω = 2.

Finding a

Using                              v² = ω²(a² – x²)

6² = 2²(a² – 10²)

⇒      a = 109

Finding ε

Substituting t = 0 in x = a sin (ωt + ε) gives

= \sqrt{109}  sin  ε

⇒          ε = 1.28 rad (see Note below).

So the equation for the motion is

x = \sqrt{109} sin (2t + 1.28).

Note
When finding ε you must be careful that you have selected the correct root of the equation.
In this case at t = 0 the particle has positive displacement and positive velocity (it is on its way out and not on its way back), so t = 0 corresponds to an angle between 0 and \frac{π}{2}.
The next root of the equation 10 = \sqrt{109} sin ε is (π – 1.28).
This lies between \frac{π}{2} and π and would be the correct value if the particle were on its way back, with displacement +10 and velocity -6.