Question 14.1: When a violin string is playing an ‘A’ with frequency 880 Hz...
When a violin string is playing an ‘A’ with frequency 880 Hz, a particle on the string oscillates with amplitude of 0.5 mm. Calculate
i) the time taken for one complete oscillation
ii) ω in terms of π and write down the equation of motion
iii) the maximum speed of the particle and its maximum acceleration
iv) the acceleration and velocity when the particle is 0.25 mm from its central position.
i) The particle does 880 oscillations per second, so the time for one oscillation is \frac{1}{880} = 0.001 13 s. This is the period of the motion.
ii) ω = 2π × frequency = 1760π
The equation of motion is \ddot{x} = – ω²x
⇒ \ddot{x} = – (1760π)²x.
iii) You can use the equations with mm so long as your units are consistent.
Maximum speed aω = 0.5 × 1760π (mms^{-1})
= 2.76 ms^{-1} (3 s.f.)
Maximum acceleration aω² = 0.5 × (1760π)² mms^{-1})
= 15 300 ms^{-2} (3 s.f.)
iv) When x = 0.25, \ddot{x} = – (1760π)² × 0.25 mms^{-2})
⇒ acceleration = -7640 ms^{-2} (3 s.f.).
The velocity is given by v² = ω²(0.5² – x²)
⇒ v = ±1760π0.25 – 0.0625(mm)s^{-1}
\begin{matrix} ⇒ velocity = ±2.39 ms^{-1} &\longleftarrow \end{matrix} \begin{matrix} \text{The particle can be travelling in either direction} \end{matrix}
When x = – 0.25 \ddot{x} = + (1760π)² × 0.25 (mms^{-2})
⇒ acceleration = +7640 ms^{-2}
The velocity is ±2.39 ms^{-1} as before.