Mathematics Extension 2

A particle is travelling in simple harmonic motion such that its velocity, in metres per second, is given by the equation

where $a,b\ne0$. What is the period of motion?

(A) $\displaystyle \frac{2b\pi}{a}$

(B) $\displaystyle \frac{2a\pi}{b}$

(C) $\displaystyle \frac{2\pi}{a}$

(D) $\displaystyle \frac{2\pi}{b}$

A particle moves in simple harmonic motion described by $\ddot x=-9(x-4)$. Find the period and the central point of motion.

A particle is moving in simple harmonic motion with centre around the origin, starting at $x=m$, where $m>0$. The displacement equation is given by $x=a\cos(nt+\alpha)$. After 1 second, the particle is at $x=r$, where $r>m$ and after another second, it returns to $x=r$.

Show that

A particle moves with simple harmonic motion with equation $x=a\sin nt$. It first returns to its starting point after $\frac\pi3$ seconds and is travelling at $15$ m/s forwards at this time. The displacement of the particle is given by

A. $x=\frac52\sin6t$

B. $x=-\frac52\sin6t$

C. $x=5\sin3t$

D. $x=-5\sin3t$

If the particle is moving at $\sqrt{15}\,\mathrm{m/s}$ when it is $2\,\mathrm m$ to the right of the centre of motion and at $2\sqrt3\,\mathrm{m/s}$ when it is $4\,\mathrm m$ to the right of the centre of motion, find the period and amplitude of the motion.

A particle is moving so that $x=2\sin3t-2\sqrt3\cos3t$. Find the initial position and velocity of the particle.

A particle is moving so that $x=2\sin3t-2\sqrt3\cos3t$. By expressing the motion in the form $A\sin(3t-\alpha)$, find the period and amplitude.

A particle is moving so that $v^2=n^2(a^2-x^2)$. Show that the particle is moving in simple harmonic motion.

The diagrams show two positions of a single piston in the cylinder chamber of a motorcycle. The piston moves vertically, in simple harmonic motion, between a maximum height of $0.17$ metres and a minimum height of $0.05$ metres.

The mass of the piston is $0.8\,\mathrm{kg}$. The piston completes $40$ cycles per second.

What is the resultant force on the piston, in newtons, that produces the maximum acceleration of the piston? Give your answer correct to the nearest newton.

For the simple harmonic motion described in part (i), find the first value of $t$ for which $x=0$, giving the answer correct to 2 decimal places.

An object is moving in simple harmonic motion along the $x$-axis. Its acceleration is $\ddot x=-4(x-3)$, where $x$ is its displacement from the origin in metres. Initially, it is $5.5$ metres to the right of the origin and moving towards the origin. Its speed is $8\,\mathrm{m\,s^{-1}}$ as it passes through the origin. Between which two values of $x$ is the particle oscillating?

A particle undergoing simple harmonic motion has a maximum acceleration of $6\ \mathrm{m/s^2}$ and a maximum velocity of $4\ \mathrm{m/s}$. What is the period of the motion?

A. $\pi$
B. $\dfrac{2\pi}{3}$
C. $3\pi$
D. $\dfrac{4\pi}{3}$

A particle is undergoing simple harmonic motion with period $\dfrac{\pi}{3}$. The central point of motion of the particle is at $x=\sqrt3$. When $t=0$ the particle has its maximum displacement of $2\sqrt3$ from the central point of motion. Find an equation for the displacement $x$ of the particle in terms of $t$.