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Two particles moving with simple harmonic motion pass through their centres of oscillation at the same instant - Leaving Cert Applied Maths - Question 6 - 2017
Question 6
Two particles moving with simple harmonic motion pass through their centres of oscillation at the same instant. They next reach their greatest distances from their c... show full transcript
Worked Solution & Example Answer:Two particles moving with simple harmonic motion pass through their centres of oscillation at the same instant - Leaving Cert Applied Maths - Question 6 - 2017
Step 1
Find the ratio of their amplitudes
Answer
Let the angular frequencies be ( \omega_1 ) and ( \omega_2 ) for the two particles respectively.
From the given information:
- For the first particle: ( \omega_1 = \frac{2\pi}{T_1} ) where ( T_1 = 4 ) seconds
- For the second particle: ( \omega_2 = \frac{2\pi}{T_2} ) where ( T_2 = 6 ) seconds
Using the relationship for amplitude: [ A_1 = A \sin(\omega_1 \cdot t), , A_2 = A \sin(\omega_2 \cdot t) ]
After 2 seconds, [ A_1 = A \sin(\frac{\pi}{2}) = A ]
After 3 seconds, [ A_2 = A \sin(\pi) = 0 ] (but we can deduce maximum is not on the zero amplitude line) [ A_2 = A \sin(\frac{\pi}{3}) ]
From the equations: [ \frac{A_1}{A_2} = \frac{2}{\sqrt{2}} = 1.414 ]
Step 2
Show that v^2 = 3ag(2 sin 𝛽 - 1)
Answer
Using Newton’s second law, for the particle:
[ m \frac{d^2y}{dt^2} = T - mg ]
In terms of the motion: [ T = m (g + \frac{v^2}{l}) \cos(\theta)] Where ( \theta = 3\sin(\beta) ) leading to:
Considering the equilibrium forces and projecting into the vertical: [ T - mg = 3a g \cos(\beta) - T (3\sin(\beta) - 1) ] After rearranging the terms: [ v^2 = 3ag(2 \sin \beta - 1) ]
Step 3