2021 A/L Physics Past Paper · Question 09 (English Medium)

Answer: (2) Graph (2)

What the question asks

We want: the graph of the wave speed $v$ against the distance $x$ along the string.

The string is made of two parts, $A$ and $B$. $B$ has twice the cross-section of $A$.

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 mass of a unit volume of a substance is termed its density.

Required equations

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

$\mu = \rho A$

$v = \sqrt{\dfrac{T}{\rho A}}$

Reasoning and calculation

What stays the same, and what changes.

The tension $T$ is the same in both parts.

The material is the same, so the density $\rho$ is the same.

The area is different: $A_B = 2A_A$.

So $\mu$ changes only because of the area.

$$\mu_B = \rho A_B = 2\rho A_A = 2\mu_A$$

The speed is:

$$v \propto \dfrac{1}{\sqrt{\mu}}$$

Now compare the two parts.

$$\dfrac{v_B}{v_A} = \sqrt{\dfrac{\mu_A}{\mu_B}} = \sqrt{\dfrac{1}{2}} = 0.71$$

So the speed in $B$ is $0.71$ of the speed in $A$.

The graph must have these features.

The speed is steady in $A$, and steady in $B$. So the graph is flat in each part.

At the joint the speed drops by a factor of $0.71$. That is a small drop, not a halving.

This is graph (2).

Reading the options

Graphs (3) and (4) rise at the joint. That cannot be, because $B$ is heavier.

Graph (5) has no change. That cannot be, because $\mu$ changes.

Graph (1) halves the speed. That needs $\mu$ four times larger.

Checks

Dimension check:

$[\sqrt{T/\mu}] = \sqrt{\dfrac{\text{kg m s}^{-2}}{\text{kg m}^{-1}}} = \sqrt{\text{m}^2\text{s}^{-2}} = \text{m s}^{-1}$

This is the dimension of a speed. It fits.

$sqrt(a / b)$

Does it make sense?

A thicker string of the same material is heavier for its length. With the same tension the wave goes slower in it.

The ratio $0.71$ is less than $1$ but more than $\tfrac{1}{2}$. That matches a small drop.

A limiting case.

If $B$ had the same area as $A$, then $\mu_B = \mu_A$ and the graph would be one flat line, graph (5).

If $B$ were four times as thick, the speed would halve, graph (1). Here the factor is two, so it is in between.

Conclusion

$v_B = \dfrac{v_A}{\sqrt{2}} = 0.71\,v_A$

$v = \sqrt{T/\mu}$ and $\mu_B = 2\mu_A$, so $v_B = 0.71\,v_A$.

The graph is flat, drops a little at the joint, then is flat again. The answer is (2).

For a string the speed is $\sqrt{T/\mu}$. A heavier string is slower.

Twice the area means twice $\mu$, so the speed falls by $\sqrt{2}$, not by $2$.

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