Speed and Acceleration in SHM | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Speed and Acceleration in SHM

Follow the oscillator from one extreme, through equilibrium, to the other extreme before using any equation.

See what changes across the motion

Swipe left or right to see the whole diagram →

The three reference positions. Displacement is measured from equilibrium. Amplitude is one centre-to-edge distance, not the full left-to-right width.

Left extreme

Speed is zero. Acceleration is greatest and points right, towards equilibrium.

Equilibrium

Speed is greatest. Acceleration is zero because displacement is zero.

Right extreme

Speed is zero. Acceleration is greatest and points left, towards equilibrium.

Speed and acceleration follow opposite patterns: one is greatest where the other is zero.

Find speed anywhere in the oscillation

The position map explains the endpoints of the equations below. At an extreme, speed must become zero. At equilibrium, speed must become its maximum value.

v=\pm\omega\sqrt{x_0^2-x^2}

Symbols

$v$ = velocity at displacement x (m s⁻¹)
$x_0$ = amplitude (m)
$x$ = displacement from equilibrium (m)
$\omega$ = angular frequency (rad s⁻¹)

v_0=\omega x_0

Symbols

$v_0$ = maximum speed (m s⁻¹)
$x_0$ = amplitude (m)

a_0=\omega^2x_0

Symbols

$a_0$ = maximum acceleration magnitude (m s⁻²)
$x_0$ = amplitude (m)

Swipe left or right to see the whole diagram →

v = ±ω√(x₀² − x²) drawn as a graph: a closed loop around the origin. It meets v = 0 exactly at x = ±x₀ (the ends of the swing) and x = 0 at v = ±v₀ (full speed through the middle).

The plus and minus signs are both needed because the object can pass the same position in either direction.
The velocity–displacement graph is a closed loop. Its four axis crossings give the two extremes and the two maximum velocities.

Worked example

Read an equation before calculating

The velocity of a block is given by the following relation.

v=0.56\cos(16t)

Find the period and amplitude.

Read the maximum speed and angular frequency from the given equation.
$v_0=0.56\ \mathrm{m\,s^{-1}}$
$\omega=16\ \mathrm{rad\,s^{-1}}$
Find the period.
$T=\dfrac{2\pi}{16}=0.39\ \mathrm{s}$
Use the maximum-speed relation.
$x_0=\dfrac{0.56}{16}=0.035\ \mathrm{m}$

Answer

T=0.39\ \mathrm{s},\qquad x_0=0.035\ \mathrm{m}

The same data and graph skill were assessed in (9702/42/M/J/25 Q5(b)).

Your turn9702/42/F/M/24 Q3(b)

A platform moves vertically with angular frequency 39 rad s⁻¹. Find the largest amplitude for which an object stays in contact.

Contact is just lost when the maximum downward acceleration equals free-fall acceleration.

Show worked answer

a_0=\omega^2x_0

x_0=\dfrac{g}{\omega^2}

x_0=\dfrac{9.81}{39^2}

x_0=6.4\times10^{-3}\ \mathrm{m}

Answer

x_0=6.4\times10^{-3}\ \mathrm{m}

Speed and Acceleration in SHM

Follow the oscillator from one extreme, through equilibrium, to the other extreme before using any equation.

See what changes across the motion

Swipe left or right to see the whole diagram →

The three reference positions. Displacement is measured from equilibrium. Amplitude is one centre-to-edge distance, not the full left-to-right width.

Left extreme

Speed is zero. Acceleration is greatest and points right, towards equilibrium.

Equilibrium

Speed is greatest. Acceleration is zero because displacement is zero.

Right extreme

Speed is zero. Acceleration is greatest and points left, towards equilibrium.

Speed and acceleration follow opposite patterns: one is greatest where the other is zero.

Find speed anywhere in the oscillation

The position map explains the endpoints of the equations below. At an extreme, speed must become zero. At equilibrium, speed must become its maximum value.

v=\pm\omega\sqrt{x_0^2-x^2}

Symbols

$v$ = velocity at displacement x (m s⁻¹)
$x_0$ = amplitude (m)
$x$ = displacement from equilibrium (m)
$\omega$ = angular frequency (rad s⁻¹)

v_0=\omega x_0

Symbols

$v_0$ = maximum speed (m s⁻¹)
$x_0$ = amplitude (m)

a_0=\omega^2x_0

Symbols

$a_0$ = maximum acceleration magnitude (m s⁻²)
$x_0$ = amplitude (m)

Swipe left or right to see the whole diagram →

v = ±ω√(x₀² − x²) drawn as a graph: a closed loop around the origin. It meets v = 0 exactly at x = ±x₀ (the ends of the swing) and x = 0 at v = ±v₀ (full speed through the middle).

The plus and minus signs are both needed because the object can pass the same position in either direction.
The velocity–displacement graph is a closed loop. Its four axis crossings give the two extremes and the two maximum velocities.

Worked example

Read an equation before calculating

The velocity of a block is given by the following relation.

v=0.56\cos(16t)

Find the period and amplitude.

Read the maximum speed and angular frequency from the given equation.
$v_0=0.56\ \mathrm{m\,s^{-1}}$
$\omega=16\ \mathrm{rad\,s^{-1}}$
Find the period.
$T=\dfrac{2\pi}{16}=0.39\ \mathrm{s}$
Use the maximum-speed relation.
$x_0=\dfrac{0.56}{16}=0.035\ \mathrm{m}$

Answer

T=0.39\ \mathrm{s},\qquad x_0=0.035\ \mathrm{m}

The same data and graph skill were assessed in (9702/42/M/J/25 Q5(b)).

Your turn9702/42/F/M/24 Q3(b)

A platform moves vertically with angular frequency 39 rad s⁻¹. Find the largest amplitude for which an object stays in contact.

Contact is just lost when the maximum downward acceleration equals free-fall acceleration.

Show worked answer

a_0=\omega^2x_0

x_0=\dfrac{g}{\omega^2}

x_0=\dfrac{9.81}{39^2}

x_0=6.4\times10^{-3}\ \mathrm{m}

Answer

x_0=6.4\times10^{-3}\ \mathrm{m}