2024 A/L Physics Past Paper · Question 19 (English Medium)

Answer: (4) $3{\cdot}0\ \text{cm}$

Given information

This question is about a pulse running up a heavy rope that hangs from a support.

A block hangs from the free end of the rope. Someone flicks the bottom of the rope, and a small pulse runs up to the top.

The rope has weight. So each part of the rope holds up everything below it: the block and the rope underneath.

What we must find: the wavelength of the pulse when it reaches the top.

How we find it: find the tension at the bottom and at the top. The speed follows the tension, and the wavelength follows the speed.

The rope is $5.0\ \text{m}$ long with mass $2.5\ \text{kg}$. The block has mass $2.0\ \text{kg}$. At the bottom the wavelength is $\lambda_b=2.0\ \text{cm}$. Take $g=10\ \text{m s}^{-2}$.

What the question asks

$T_b$, $T_t$ ($\text{N}$): tension at the bottom and at the top.

$m$ ($\text{kg m}^{-1}$): mass per unit length of the rope.

$v_b$, $v_t$ ($\text{m s}^{-1}$): speed of the pulse at the bottom and at the top.

$f$ ($\text{Hz}$): frequency.

$\lambda_b$, $\lambda_t$ ($\text{cm}$): wavelength at the bottom and at the top.

Relevant theory

If T is the tension in a string and m its linear density (mass per unit length) then the velocity v of a transverse wave along the string can be expressed as,

The relationship among the wave velocity v, frequency f and the wavelength λ, can be written in the form of v = f λ. This equation can be applied to all types of waves.

What is happening: the hand at the bottom sets the frequency. Every part of the rope swings as often as the hand, so $f$ is the same all the way up.

Going up, each part of the rope carries more weight below it, so the tension grows. A tighter rope carries the pulse faster.

Same frequency, faster speed: the pulse stretches out, and its wavelength grows.

For example, think of the climbing rope in a school gym, with a schoolbag tied to its end.

A student flicks the rope just above the bag. A ripple climbs up the rope.

Near the ceiling the rope is pulled tight by everything hanging below. The ripple speeds up there and spreads out into a longer bump.

In words: a tighter rope carries a ripple faster, and a faster ripple at the same beat is a longer one.

Required equations

$v=\sqrt{\dfrac{T}{m}}$

$v=f\lambda$

$\dfrac{\lambda_t}{\lambda_b}=\dfrac{v_t}{v_b}=\sqrt{\dfrac{T_t}{T_b}}$

Applying it to this question

$20$

$T_b=2.0\times10$

$45$

$T_t=(2.0+2.5)\times10$

$1.5$

$\dfrac{\lambda_t}{\lambda_b}=\sqrt{\dfrac{45}{20}}=\sqrt{2.25}$

$\lambda_t=2.0\times1.5$

$3.0$

$\lambda_t$

Reasoning and calculation

The tension grows $2.25$ times from bottom to top, so the speed and the wavelength grow $1.5$ times. $2.0\ \text{cm}$ becomes $3.0\ \text{cm}$, option (4).

$1.5\ \text{cm}$ makes the pulse shorter at the top. That would mean it slows down going up. But the top of the rope is pulled harder, so the pulse goes faster there and gets longer.

$2.0\ \text{cm}$ keeps the wavelength the same. That is true only if the tension is the same everywhere. Here the rope has weight, so the tension grows from $20\ \text{N}$ at the bottom to $45\ \text{N}$ at the top.

$2.5\ \text{cm}$ scales the wavelength by the masses, $2.5/2.0$. But the top holds the block and the rope together, $45\ \text{N}$, and the speed follows the square root of the tension.

$4.0\ \text{cm}$ doubles the wavelength. For that the speed must double, so the tension must be four times bigger, $80\ \text{N}$. The top tension is only $45\ \text{N}$.

Checks

$lb*sqrt(Tt/Tb)$

$m=\dfrac{2.5}{5.0}=0.5\ \text{kg m}^{-1}$

$v_b=\sqrt{\dfrac{20}{0.5}}=\sqrt{40},\quad v_t=\sqrt{\dfrac{45}{0.5}}=\sqrt{90}$

$\dfrac{v_t}{v_b}=\sqrt{\dfrac{90}{40}}=1.5$

The answer is a length in $\text{cm}$, as a wavelength should be. It is longer than at the bottom, because the top is pulled harder.

$\sqrt{T_t/T_b}$ is $\sqrt{\text{N}/\text{N}}$, a pure number, so $\lambda_t$ keeps the unit of $\lambda_b$, $\text{cm}$.

Conclusion

$\lambda_t=3.0\ \text{cm}$

At the top the wavelength is $3.0\ \text{cm}$. The answer is (4).