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Waves and Oscillations

B.Sc. (Hons) Physics

Waves and Oscillations

Delhi University · B.Sc. (Hons) Physics · Semester I · DSC-3

Unit-wise notes in simple English with diagrams, and every previous-year question from papers 1069, 1403, 1631 and 5989 solved (57 sub-parts).

Start here

How to use these notes

Pick a unit below.
Read the concept notes and look at the diagrams.
At the end of each unit, compare the previous-year solutions with your own working.

12 formulas you will use again and again

Topic	Formula	Meaning
SHM	$\ddot x+\omega_0^2x=0$	Acceleration points toward the mean position
Spring–mass	$\omega_0=\sqrt{k/m}$ , $T=2\pi\sqrt{m/k}$	Larger stiffness = faster; larger mass = slower
Spring energy	$E=\tfrac12kA^2=\tfrac12m\omega_0^2A^2$	Constant total energy in undamped SHM
Damping	$m\ddot x+b\dot x+kx=0$	Opposing friction force is $-b\dot x$
Damped frequency	$\omega_d=\sqrt{\omega_0^2-\beta^2}$	Valid in underdamping
Amplitude decay	$A(t)=A_0e^{-\beta t}$	Amplitude reduces exponentially
Log decrement	$\delta=\ln(A_n/A_{n+1})=\beta T_d$	Loss of amplitude per cycle
Quality factor	$Q\simeq\omega_0/(2\beta)$	Sharpness of resonance, weak damping
Forced amplitude	$X=F_0/\sqrt{(k-m\omega^2)^2+(b\omega)^2}$	Response at driving angular frequency $\omega$
Beats	$f_b=\vert f_1-f_2\vert$	Number of loud–soft cycles per second
String modes	$f_n=\tfrac{n}{2L}\sqrt{T/\mu}$	Fixed ends; $\mu$ = mass per length
Travelling wave	$\partial_t^2 y=v^2\partial_x^2y$	Wave speed $v=\sqrt{T/\mu}$ on a string

Last-minute revision

SHM: $\ddot x+\omega_0^2x=0$ , $T=2\pi/\omega_0$ .
Damping: $\beta=b/(2m)$ , $\omega_d=\sqrt{\omega_0^2-\beta^2}$ , $A=A_0e^{-\beta t}$ .
Decay questions: $\delta=(1/N)\ln(A_0/A_N)$ , $\beta=\delta/T$ , $\tau_A=1/\beta$ .
LCR: $\omega_0=1/\sqrt{LC}$ , $Q=\omega_0L/R$ , $\Delta f=R/(2\pi L)$ .
Beats: $f_b=\vert f_1-f_2\vert$ .
Springs: series $k_{eq}=k_1k_2/(k_1+k_2)$ ; working together $k_{eq}=k_1+k_2$ .
String: $f_n=n\sqrt{T/\mu}/(2L)$ .
Normal modes: assume a common frequency, solve the small-displacement equations, then find each mode's amplitude pattern.
Physical pendulum: $T=2\pi\sqrt{I_p/(Mgh)}$ .
Proof of power balance: Multiply $m\ddot x+b\dot x+kx=F$ by $\dot x$ and average over one cycle.

Where each printed diagram is

All nine printed figure locations are redrawn as diagrams inside the solution that uses them.

Paper	Question	Diagram	Unit
1069	Q1(b), Fig. 1	Mass between two springs	Unit 1
1069	Q5(a), Fig. 2	String of length 3a with two masses	Unit 3
1403	Q4(b), Fig. 1	Two masses, three springs	Unit 3
1631	Q1(e), Fig. 1	Mass on a string, transverse oscillation	Unit 1
1631	Q2(b), Fig. 2(a), 2(b)	Springs in series, mass between two springs	Unit 1
1631	Q5(a), Fig. 3	Two vertically coupled masses	Unit 3
5989	Q3(c), (a) (b) (c)	Three spring arrangements	Unit 1
5989	Q5(b), unnumbered	Two masses coupled by a spring	Unit 3

Unit 1 · Simple Harmonic Motion

12 hours · 26 solved previous-year questions

The big picture

SHM is the building block of the whole course. A restoring force pulls a body back to the middle, and the result is a smooth cosine. Add several of these together and you get beats, circles and Lissajous patterns.

Rendering diagram…

1. The SHM equation

Definition
If the acceleration is proportional to the displacement and always points to the middle, the motion is simple harmonic.

\ddot x+\omega_0^2x=0 \Rightarrow x=A\cos(\omega_0t+\phi)

$\omega_0=2\pi f=2\pi/T$ . $A$ and $\phi$ come from the starting conditions.

Quantity	Formula	Biggest value	Where
Displacement	$x=A\cos(\omega t+\phi)$	$A$	ends of the motion
Velocity	$v=-\omega A\sin(\omega t+\phi)$	$\omega A$	middle position
Acceleration	$a=-\omega^2x$	$\omega^2A$	ends of the motion
Speed at $x$	$v=\omega\sqrt{A^2-x^2}$

Figure 1. Velocity is a quarter cycle ahead of displacement. Acceleration is the opposite of displacement.

Energy in SHM

K=\tfrac12mv^2,\quad U=\tfrac12kx^2,\quad E=K+U=\tfrac12kA^2=\tfrac12m\omega^2A^2

Figure 2. The two parabolas always add up to the flat line E.

Exam tip
Exam trick. Average kinetic energy and average potential energy over a cycle are equal, each $E/2$ . They are equal at the instant $x=\pm A/\sqrt2$ .

Linearity and superposition

The SHM equation is linear and homogeneous. So if $x_1$ and $x_2$ are solutions, then $c_1x_1+c_2x_2$ is also a solution. This is the principle of superposition. It fails for non-linear equations such as the large-angle pendulum.

2. Rotating vector picture

Draw an arrow of length $A$ that turns anticlockwise at angular speed $\omega$ . Its shadow on a vertical line moves in SHM. One full turn of the arrow equals one full oscillation.

Figure 3. The projection of a uniformly rotating arrow is simple harmonic.

This picture makes adding oscillations easy. Adding two SHMs of the same frequency means adding two arrows head to tail.

3. Real systems that do SHM

Figure 4. Simple pendulum and compound pendulum.

System	Period	Remember
Mass on a spring	$T=2\pi\sqrt{m/k}$	Gravity only shifts the middle position
Spring with its own mass $m_s$	$T=2\pi\sqrt{(M+m_s/3)/K}$	One third of the spring mass moves with the load
Simple pendulum	$T=2\pi\sqrt{l/g}$	Small angles only
Compound pendulum	$T=2\pi\sqrt{(k_g^2+h^2)/(gh)}$	Equivalent length $l=(k_g^2+h^2)/h$

Compound (bar) pendulum facts

$O$ is the pivot, $G$ the centre of gravity, $C$ the centre of oscillation. Pivot and $C$ can be swapped without changing the period (they are conjugate).
$h_1h_2=k_g^2$ and $h_1+h_2=l$ for the two conjugate points on the same side of $G$ .
The period is least when $h=k_g$ .

Important
Kater's pendulum. A bar with two knife edges at distances $l_1$ and $l_2$ from $G$ , on opposite sides. Time it about each edge to get $T_1$ and $T_2$ :

\dfrac{8\pi^2}{g}=\dfrac{T_1^2+T_2^2}{l_1+l_2}+\dfrac{T_1^2-T_2^2}{l_1-l_2}

If $T_1\approx T_2$ , the second term almost vanishes and $g=4\pi^2(l_1+l_2)/T^2$ . That is why the edges are adjusted until the periods match.

Spring combinations

Figure 5. Series is softer, side-by-side is stiffer.

A uniform spring cut into pieces: stiffness is inversely proportional to length, $k\propto 1/l$ . A piece that is $1/n$ of the length has constant $nk$ .
Series springs: same force, extensions add, so $1/k=1/k_1+1/k_2$ .
Parallel or on both sides of a mass: same extension, forces add, so $k=k_1+k_2$ .

4. Adding oscillations along one line

N equal oscillations with a fixed phase step

A=A_0\left\vert \dfrac{\sin(N\phi/2)}{\sin(\phi/2)}\right\vert ,\qquad \text{resultant phase}=\dfrac{(N-1)\phi}{2}

$\phi\to0$ gives $NA_0$ . The resultant is zero when $N\phi=2\pi,4\pi,\dots$ but $\phi\ne 0,2\pi,\dots$

Beats

Two frequencies that are almost equal give a loud-soft pattern. The beat frequency is the difference.

x=2a\cos\!\Big(\dfrac{\omega_1-\omega_2}{2}t\Big)\cos\!\Big(\dfrac{\omega_1+\omega_2}{2}t\Big),\qquad f_{beat}=\vert f_1-f_2\vert

Figure 6. Fast wiggles inside a slow envelope. With unequal amplitudes the minimum is |a₁-a₂|, not zero.

5. Adding oscillations at right angles

x=a\sin\omega t,\; y=b\sin(\omega t+\varphi) \Rightarrow \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{2xy}{ab}\cos\varphi=\sin^2\varphi

Phase $\varphi$	Path
$0$	Straight line $y=(b/a)x$
$\pi$	Straight line $y=-(b/a)x$
$\pi/2$ , $a\ne b$	Ellipse with axes along $x,y$
$\pi/2$ , $a=b$	Circle
Anything else	Tilted ellipse

Lissajous figures (different frequencies)

Exam tip
How to read the ratio. Draw a horizontal line and a vertical line that just touch the figure. Then $f_x : f_y$ = (number of touches on a vertical side) : (number of touches on a horizontal side).

Opening your notes…

Just a moment

Waves and Oscillations

B.Sc. (Hons) Physics

Waves and Oscillations

Delhi University · B.Sc. (Hons) Physics · Semester I · DSC-3

Unit-wise notes in simple English with diagrams, and every previous-year question from papers 1069, 1403, 1631 and 5989 solved (57 sub-parts).

Start here

How to use these notes

Pick a unit below.
Read the concept notes and look at the diagrams.
At the end of each unit, compare the previous-year solutions with your own working.

12 formulas you will use again and again

Topic	Formula	Meaning
SHM	$\ddot x+\omega_0^2x=0$	Acceleration points toward the mean position
Spring–mass	$\omega_0=\sqrt{k/m}$ , $T=2\pi\sqrt{m/k}$	Larger stiffness = faster; larger mass = slower
Spring energy	$E=\tfrac12kA^2=\tfrac12m\omega_0^2A^2$	Constant total energy in undamped SHM
Damping	$m\ddot x+b\dot x+kx=0$	Opposing friction force is $-b\dot x$
Damped frequency	$\omega_d=\sqrt{\omega_0^2-\beta^2}$	Valid in underdamping
Amplitude decay	$A(t)=A_0e^{-\beta t}$	Amplitude reduces exponentially
Log decrement	$\delta=\ln(A_n/A_{n+1})=\beta T_d$	Loss of amplitude per cycle
Quality factor	$Q\simeq\omega_0/(2\beta)$	Sharpness of resonance, weak damping
Forced amplitude	$X=F_0/\sqrt{(k-m\omega^2)^2+(b\omega)^2}$	Response at driving angular frequency $\omega$
Beats	$f_b=\vert f_1-f_2\vert$	Number of loud–soft cycles per second
String modes	$f_n=\tfrac{n}{2L}\sqrt{T/\mu}$	Fixed ends; $\mu$ = mass per length
Travelling wave	$\partial_t^2 y=v^2\partial_x^2y$	Wave speed $v=\sqrt{T/\mu}$ on a string

Last-minute revision

SHM: $\ddot x+\omega_0^2x=0$ , $T=2\pi/\omega_0$ .
Damping: $\beta=b/(2m)$ , $\omega_d=\sqrt{\omega_0^2-\beta^2}$ , $A=A_0e^{-\beta t}$ .
Decay questions: $\delta=(1/N)\ln(A_0/A_N)$ , $\beta=\delta/T$ , $\tau_A=1/\beta$ .
LCR: $\omega_0=1/\sqrt{LC}$ , $Q=\omega_0L/R$ , $\Delta f=R/(2\pi L)$ .
Beats: $f_b=\vert f_1-f_2\vert$ .
Springs: series $k_{eq}=k_1k_2/(k_1+k_2)$ ; working together $k_{eq}=k_1+k_2$ .
String: $f_n=n\sqrt{T/\mu}/(2L)$ .
Normal modes: assume a common frequency, solve the small-displacement equations, then find each mode's amplitude pattern.
Physical pendulum: $T=2\pi\sqrt{I_p/(Mgh)}$ .
Proof of power balance: Multiply $m\ddot x+b\dot x+kx=F$ by $\dot x$ and average over one cycle.

Where each printed diagram is

All nine printed figure locations are redrawn as diagrams inside the solution that uses them.

Paper	Question	Diagram	Unit
1069	Q1(b), Fig. 1	Mass between two springs	Unit 1
1069	Q5(a), Fig. 2	String of length 3a with two masses	Unit 3
1403	Q4(b), Fig. 1	Two masses, three springs	Unit 3
1631	Q1(e), Fig. 1	Mass on a string, transverse oscillation	Unit 1
1631	Q2(b), Fig. 2(a), 2(b)	Springs in series, mass between two springs	Unit 1
1631	Q5(a), Fig. 3	Two vertically coupled masses	Unit 3
5989	Q3(c), (a) (b) (c)	Three spring arrangements	Unit 1
5989	Q5(b), unnumbered	Two masses coupled by a spring	Unit 3

Unit 1 · Simple Harmonic Motion

12 hours · 26 solved previous-year questions

The big picture

Rendering diagram…

1. The SHM equation

Definition
If the acceleration is proportional to the displacement and always points to the middle, the motion is simple harmonic.

\ddot x+\omega_0^2x=0 \Rightarrow x=A\cos(\omega_0t+\phi)

$\omega_0=2\pi f=2\pi/T$ . $A$ and $\phi$ come from the starting conditions.

Quantity	Formula	Biggest value	Where
Displacement	$x=A\cos(\omega t+\phi)$	$A$	ends of the motion
Velocity	$v=-\omega A\sin(\omega t+\phi)$	$\omega A$	middle position
Acceleration	$a=-\omega^2x$	$\omega^2A$	ends of the motion
Speed at $x$	$v=\omega\sqrt{A^2-x^2}$

Figure 1. Velocity is a quarter cycle ahead of displacement. Acceleration is the opposite of displacement.

Energy in SHM

K=\tfrac12mv^2,\quad U=\tfrac12kx^2,\quad E=K+U=\tfrac12kA^2=\tfrac12m\omega^2A^2

Figure 2. The two parabolas always add up to the flat line E.

Exam tip
Exam trick. Average kinetic energy and average potential energy over a cycle are equal, each $E/2$ . They are equal at the instant $x=\pm A/\sqrt2$ .

Linearity and superposition

2. Rotating vector picture

Draw an arrow of length $A$ that turns anticlockwise at angular speed $\omega$ . Its shadow on a vertical line moves in SHM. One full turn of the arrow equals one full oscillation.

Figure 3. The projection of a uniformly rotating arrow is simple harmonic.

This picture makes adding oscillations easy. Adding two SHMs of the same frequency means adding two arrows head to tail.

3. Real systems that do SHM

Figure 4. Simple pendulum and compound pendulum.

System	Period	Remember
Mass on a spring	$T=2\pi\sqrt{m/k}$	Gravity only shifts the middle position
Spring with its own mass $m_s$	$T=2\pi\sqrt{(M+m_s/3)/K}$	One third of the spring mass moves with the load
Simple pendulum	$T=2\pi\sqrt{l/g}$	Small angles only
Compound pendulum	$T=2\pi\sqrt{(k_g^2+h^2)/(gh)}$	Equivalent length $l=(k_g^2+h^2)/h$

Compound (bar) pendulum facts

$O$ is the pivot, $G$ the centre of gravity, $C$ the centre of oscillation. Pivot and $C$ can be swapped without changing the period (they are conjugate).
$h_1h_2=k_g^2$ and $h_1+h_2=l$ for the two conjugate points on the same side of $G$ .
The period is least when $h=k_g$ .

Important
Kater's pendulum. A bar with two knife edges at distances $l_1$ and $l_2$ from $G$ , on opposite sides. Time it about each edge to get $T_1$ and $T_2$ :

\dfrac{8\pi^2}{g}=\dfrac{T_1^2+T_2^2}{l_1+l_2}+\dfrac{T_1^2-T_2^2}{l_1-l_2}

If $T_1\approx T_2$ , the second term almost vanishes and $g=4\pi^2(l_1+l_2)/T^2$ . That is why the edges are adjusted until the periods match.

Spring combinations

Figure 5. Series is softer, side-by-side is stiffer.

A uniform spring cut into pieces: stiffness is inversely proportional to length, $k\propto 1/l$ . A piece that is $1/n$ of the length has constant $nk$ .
Series springs: same force, extensions add, so $1/k=1/k_1+1/k_2$ .
Parallel or on both sides of a mass: same extension, forces add, so $k=k_1+k_2$ .

4. Adding oscillations along one line

N equal oscillations with a fixed phase step

A=A_0\left\vert \dfrac{\sin(N\phi/2)}{\sin(\phi/2)}\right\vert ,\qquad \text{resultant phase}=\dfrac{(N-1)\phi}{2}

$\phi\to0$ gives $NA_0$ . The resultant is zero when $N\phi=2\pi,4\pi,\dots$ but $\phi\ne 0,2\pi,\dots$

Beats

Two frequencies that are almost equal give a loud-soft pattern. The beat frequency is the difference.

x=2a\cos\!\Big(\dfrac{\omega_1-\omega_2}{2}t\Big)\cos\!\Big(\dfrac{\omega_1+\omega_2}{2}t\Big),\qquad f_{beat}=\vert f_1-f_2\vert

Figure 6. Fast wiggles inside a slow envelope. With unequal amplitudes the minimum is |a₁-a₂|, not zero.

5. Adding oscillations at right angles

x=a\sin\omega t,\; y=b\sin(\omega t+\varphi) \Rightarrow \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{2xy}{ab}\cos\varphi=\sin^2\varphi

Phase $\varphi$	Path
$0$	Straight line $y=(b/a)x$
$\pi$	Straight line $y=-(b/a)x$
$\pi/2$ , $a\ne b$	Ellipse with axes along $x,y$
$\pi/2$ , $a=b$	Circle
Anything else	Tilted ellipse

Lissajous figures (different frequencies)

Exam tip
How to read the ratio. Draw a horizontal line and a vertical line that just touch the figure. Then $f_x : f_y$ = (number of touches on a vertical side) : (number of touches on a horizontal side).