CAIE A-Level Physics 17.1 Simple Harmonic Motion

Practise defining SHM, interpreting phase and parameters, applying displacement, velocity and acceleration equations and analysing their graphs.

Exam points

interpret SHM parameters, equations and phase relationships
apply a = −ω²x, x = x₀ sinωt and the SHM velocity relationships
analyse displacement, velocity and acceleration graphs in SHM

Covered syllabus points

17.1.1—The terms displacement, amplitude, period, frequency, angular frequency and 17.1.2—Simple harmonic motion occurs when acceleration is proportional to 17.1.3—A = –ω2x and recall and use, as a solution to this equation, x = x0 sin ωt 17.1.4—The equations v = v0 cos ωt and v = ± ω ()xx0 22−17.1.5—Graphical representations of the variations of displacement, velocity and

[Maximum number: 9]

A spring is hung vertically from a fixed point. A mass M is hung from the other end of the spring, as illustrated in Fig. 3.1.

Fig. 3.1

The mass is displaced downwards and then released. The subsequent motion of the mass is simple harmonic.

The variation with time t of the length L of the spring is shown in Fig. 3.2.

Fig. 3.2

(a)

State:

[ 1 ]

(i)

one time at which the mass is moving with maximum speed
time =

[ 1 ]

(b)

Use data from Fig. 3.2 to determine, for the motion of the mass:

[ 6 ]

(i)

the angular frequency $\omega$ $\omega=$ $\mathrm{rads}^{-1}$

[ 2 ]

C1

$\begin{aligned} \omega =2 \pi / T =2 \pi / 0.40 =16 \mathrm{rad} \mathrm{~s}^{-1} \end{aligned}$

A1

(ii)

the maximum speed
maximum speed = $\mathrm{ms}^{-1}$

[ 2 ]

$v_{0}=\omega \mathrm{X}_{0}$

C1

$\begin{aligned} =15.7 \times 2.5 \times 10^{-2} =0.39 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}$

A1

or

tangent drawn at steepest part and working to show attempted calculation of gradient

(C1)

leading to $v_{0}=0.39 \mathrm{~m} \mathrm{~s}^{-1}$ (allow ± $0.15 \mathrm{~m} \mathrm{~s}^{-1}$ )

(A1)

(iii)

the magnitude of the maximum acceleration.
maximum acceleration = $\mathrm{ms}^{-2}$

[ 2 ]

$a_{0}=\omega^{2} x_{0}$

C1

$\begin{aligned} a_{0} =\left(15.7^{2} \times 2.5 \times 10^{-2}\right) =6.2 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$

A1

or

$a_{0}=\omega V_{0}$

(C1)

$\begin{aligned} a_{0} =15.7 \times 0.39 =6.2 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$

(A1)

(c)

The mass M is now suspended from two springs, each identical to that in Fig. 3.1, as shown in Fig. 3.3.

Fig. 3.3

Suggest and explain the change, if any, in the period of oscillation of the mass. A numerical answer is not required.

[ 2 ]

[Maximum number: 4]

A small wooden block (cuboid) of mass m floats in water, as shown in Fig. 3.1.

Fig. 3.1

The top face of the block is horizontal and has area A. The density of the water is $\rho$ .

(a)

The block in (b) is now released so that it oscillates vertically.

The resultant force F acting on the block is given by

F=-A g \rho x

where g is the gravitational field strength and x is the vertical displacement of the block from the equilibrium position.

[ 4 ]

(i)

Explain why the oscillations of the block are simple harmonic.

[ 2 ]

A, g and $\rho$ all constant so $F \propto x$

B1

minus sign means F and x are in opposite directions

B1

(ii)

Show that the angular frequency $\omega$ of the oscillations is given by

\omega=\sqrt{\frac{A \rho g}{m}} .

[ 2 ]

$\left(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}\right.$ so $) \mathrm{a}=(-) \frac{\text { Agpx }}{\mathrm{m}}$

M1

so $\omega^{2}=\frac{\mathrm{Ag} \rho}{\mathrm{m}}$ hence $\omega=\sqrt{\frac{\mathrm{Ag} \rho}{\mathrm{m}}}$

A1

All question bank results loaded

Covered syllabus points

17.1 Simple harmonic oscillations question 1

Question (a)

Question (i)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question (c)

17.1 Simple harmonic oscillations question 2

Question (a)

Question (i)

Question (ii)