"Oscillations and Waves" — Practice

Question 1

A particle in SHM has amplitude $A$. At what displacement are its kinetic and potential energies equal?

Accepted Answer

$U=	frac12m\omega^{2}x^{2}$ and $K=	frac12m\omega^{2}(A^{2}-x^{2})$.

Setting them equal: $x^{2}=A^{2}-x^{2}\Rightarrow 2x^{2}=A^{2}$

$x=\mathbf{\dfrac{A}{\sqrt2}}\approx0.71A$

**Not $A/2$.** Energy is quadratic in displacement, so half the energy sits at $1/\sqrt2$ of the amplitude.

Question 2

State the phase relationships between displacement, velocity and acceleration in SHM, and where each is maximum.

Accepted Answer

| Quantity | Maximum | Where | Phase vs $x$ |
|---|---|---|---|
| $x$ | $A$ | extremes | — |
| $v$ | $A\omega$ | **equilibrium** | leads by $\pi/2$ |
| $a$ | $A\omega^{2}$ | **extremes** | antiphase, $\pi$ |

At the extremes the particle is momentarily at rest but under maximum restoring force. At equilibrium it moves fastest but feels no force at all.

The antiphase relation is just $a=-\omega^{2}x$ read as a statement about phase.

Question 3

Why is sound in air a longitudinal wave rather than a transverse one?

Accepted Answer

Restoring a **sideways** displacement is a **shear**, and resisting shear requires a non-zero rigidity modulus.

Air, like every fluid, has **zero rigidity modulus** — apply a tangential stress and it simply flows. So a transverse mechanical wave cannot propagate through it, and only longitudinal (compression-rarefaction) waves survive.

Seismologists use exactly this: S-waves, which are transverse, fail to cross the Earth's outer core. That is how we know the outer core is liquid.

Question 4

A mass hangs from two identical springs of stiffness $k$, first in parallel and then in series. Compare the periods.

Accepted Answer

**Parallel:** $k_{eff}=2k$.  **Series:** $k_{eff}=k/2$.

The series arrangement is **four times softer**.

Since $T=2\pi\sqrt{m/k}\propto1/\sqrt{k}$:

$\dfrac{T_{series}}{T_{parallel}}=\sqrt{\dfrac{2k}{k/2}}=\mathbf{2}$

The series period is **twice** the parallel one. Note that series stiffness always falls below both individual values, and parallel always rises above both.

Question 5

A 1 m pendulum is in a lift. Find its period when the lift accelerates upward at $2\ 	ext{m/s}^2$, and when it is in free fall. Take $g=10$.

Accepted Answer

**Accelerating up:** $g_{eff}=g+a=12$

$T=2\pi\sqrt{\dfrac{1}{12}}\approx\mathbf{1.81\ 	ext{s}}$

Shorter than the $2\pi\sqrt{1/10}\approx1.99$ s at rest — a stronger effective gravity means a faster swing.

**Free fall:** $g_{eff}=g-g=0$, so $T	o\infty$.

**The pendulum stops oscillating entirely.** With no effective gravity there is no restoring force, and the bob simply floats wherever it was left.

Question 6

A 500 Hz wave travels at $350\ 	ext{m/s}$. Find the phase difference between two points $0.175\ 	ext{m}$ apart, and at one point after $0.5\ 	ext{ms}$.

Accepted Answer

$\lambda=\dfrac{350}{500}=0.7\ 	ext{m}$ and $T=\dfrac{1}{500}=2\ 	ext{ms}$.

**Spatial:** $\Delta\phi=\dfrac{2\pi}{\lambda}\Delta x=\dfrac{2\pi}{0.7}(0.175)=\mathbf{\dfrac{\pi}{2}}$

**Temporal:** $\Delta\phi=\dfrac{2\pi}{T}\Delta t=\dfrac{2\pi}{2}(0.5)=\mathbf{\dfrac{\pi}{2}}$

Both are a quarter cycle. The two points are $\lambda/4$ apart, and the interval is $T/4$ — the same fraction in each case, which is why the phase shift matches.

Question 7

An open pipe and a closed pipe sound the same fundamental. Compare their lengths and list the first three harmonics of each.

Accepted Answer

$\dfrac{v}{2L_{open}}=\dfrac{v}{4L_{closed}}\Rightarrow L_{closed}=\dfrac{L_{open}}{2}$

The closed pipe is **half as long** for the same note.

**Open:** $f$, $2f$, $3f$ — all harmonics.

**Closed:** $f$, $3f$, $5f$ — **odd only**.

Same pitch, completely different timbre. The closed pipe's missing even harmonics are exactly what give it a hollow tone.

Question 8

A tuning fork gives 5 beats per second with a 256 Hz fork. Loaded with wax, it gives 3 beats per second. Find its frequency.

Accepted Answer

Five beats $\Rightarrow$ the fork is at **251 or 261 Hz**.

Wax **lowers** the loaded fork's frequency.

- If it were 251, lowering it moves it **further** from 256 and the beat rate would **rise**.
- If it were 261, lowering it moves it **closer** and the beat rate **falls**.

The rate fell from 5 to 3, so $f=\mathbf{261\ 	ext{Hz}}$.

The beat count alone could never settle this — only the *direction* of the change does.

Question 9

Explain why a standing wave carries no net energy, and how the phase relationship between particles differs from a progressive wave.

Accepted Answer

**No net energy.** A standing wave is the superposition of two identical waves travelling in **opposite** directions. Each carries equal energy, so the two transports cancel and the net flow is zero. Energy sloshes between kinetic and potential locally, but goes nowhere.

**Phase.** In a **progressive** wave each particle lags the one before it by $\dfrac{2\pi}{\lambda}\Delta x$ — a smooth, continuous phase gradient along the direction of travel.

In a **standing** wave there is no gradient at all. Every particle between two adjacent nodes is **exactly in phase**; particles on opposite sides of a node are **exactly antiphase**. Only the amplitude varies with position, as $2A\sin kx$.

That separation of $x$ and $t$ in $y=2A\sin(kx)\cos(\omega t)$ is the mathematical statement of both facts at once.

Question 10

Newton computed the speed of sound in air as $\sqrt{P/\rho}$ and got a value about 15% below measurement. Explain the error and Laplace's correction.

Accepted Answer

**Newton's assumption:** the compressions and rarefactions are **isothermal** — slow enough for heat to flow and keep the temperature uniform. That gives the isothermal bulk modulus, which is simply $P$, hence $v=\sqrt{P/\rho}$.

**Laplace's correction:** sound oscillations are far too rapid for heat to flow between compressions and rarefactions, so they are **adiabatic**. The relevant bulk modulus is $\gamma P$:

$$v=\sqrt{\frac{\gamma P}{\rho}}$$

For air $\gamma=1.4$, so the corrected speed is larger by $\sqrt{1.4}\approx1.18$ — an 18% increase, which closes Newton's 15% shortfall.

Note the consequence: since $P/\rho$ is fixed at a given temperature by the gas law, sound speed depends on **temperature alone**, not on pressure.

Question 11

A particle moves according to $x=3\sin 2t+4\cos 2t$ (SI units). Show that this is simple harmonic motion, and find its amplitude and period.

Accepted Answer

Write $x=R\sin(2t+\phi)$ with $R=\sqrt{3^{2}+4^{2}}$:

$$x=5\sin(2t+\phi),\qquad 	an\phi=\dfrac43$$

A **single** sinusoid, so differentiating twice gives $a=-4x$ — of the form $a=-\omega^{2}x$, which is the definition. It is SHM.

$A=\mathbf{5\ 	ext{m}}, \qquad \omega=2\ 	ext{rad/s}, \qquad T=\dfrac{2\pi}{2}=\mathbf{\pi\ 	ext{s}}$

*Why the combination works here but not always:* both terms share the **same** $\omega$, so they collapse to one sinusoid. Had the second term been $4\cos 4t$, the two would not combine, $a$ would not be proportional to $x$, and the motion would be periodic but not simple harmonic.

Question 12

In a resonance tube experiment with a $512\ 	ext{Hz}$ fork, the first resonance occurs at an air column of $16.0\ 	ext{cm}$ and the second at $49.5\ 	ext{cm}$. Find the speed of sound and the end correction.

Accepted Answer

A resonance tube is a **closed** pipe, so resonances occur at $\dfrac{\lambda}{4}$ and $\dfrac{3\lambda}{4}$, each measured from the effective end:

$l_1+e=\dfrac{\lambda}{4}, \qquad l_2+e=\dfrac{3\lambda}{4}$

**Subtracting eliminates $e$** — which is the whole design of the experiment:

$l_2-l_1=\dfrac{\lambda}{2} \Rightarrow \lambda=2(49.5-16.0)=67.0\ 	ext{cm}$

$v=f\lambda=512(0.670)=\mathbf{343\ 	ext{m/s}}$

Now recover the end correction from the first resonance:

$e=\dfrac{\lambda}{4}-l_1=16.75-16.0=\mathbf{0.75\ 	ext{cm}}$

*Sanity check:* $343\ 	ext{m/s}$ is the speed of sound at about $20°	ext{C}$, which is a reasonable laboratory temperature. And $e\approx0.6r$ implies a tube radius of about $1.25$ cm, a normal size.

Practice — "Oscillations and Waves"

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