Standing wave in a tube -- multiple choice question

[EF17xx]

Homework Statement

The correct answer according to answer sheet is answer B.

Homework Equations

λn= 2L/n (both ends open)
λn = 4L/n (one end open one end closed)
c=λf

Speed of sound does not change. as it is still in air... (correct assumption? )

The Attempt at a Solution

f1= 500Hz
L = λ/2
f= v/2L

Tube cut in half means L/2 = λ/4
L(new)= λ/4
f= v/4L(new)

Lnew = L/2

Therefore

v/4(L/2) = f
f= v/2L
therefore f is still 500Hz

I am just wondering if my reasoning and assumptions are CORRECT could someone please confirm.

 

[mjc123]

Yes - half-wavelength in the open tube, quarter-wavelength in the closed tube half as long - same wavelength and frequency.

 

[tnich]

Homework Statement

View attachment 222833
The correct answer according to answer sheet is answer B.

Homework Equations

λn= 2L/n (both ends open)
λn = 4L/n (one end open one end closed)
c=λf

Speed of sound does not change. as it is still in air... (correct assumption? )

The Attempt at a Solution

f1= 500Hz
L = λ/2
f= v/2L

Tube cut in half means L/2 = λ/4
L(new)= λ/4
f= v/4L(new)

Lnew = L/2

Therefore

v/4(L/2) = f
f= v/2L
therefore f is still 500Hz

I am just wondering if my reasoning and assumptions are CORRECT could someone please confirm.

It appears that you have cut the tube in half twice and somehow ended up with the same frequency at the end. I think you might start by asking yourself, "What does cutting the tube in half do to the frequency?" and "What does closing the end do to the frequency?"

Part of what makes your reasoning difficult to follow is notation like v/4(L/2) = f. Does that mean $\frac {vL}8$ or $\frac v{2L}$? Also, you have used variables Lnew and L(new). Do they mean the same thing or something different? Using clearer notation can help you and other people understand what you wrote. Take a look at the LaTeX guide.

 

[EF17xx]

Lnew is the length of the tube after it has been cut in half expressed in terms of the original length of the tube (L) therefore L(new) = L/2

(v) / (4(L/2))

Therefore f= v/2L

 

[tnich]

Lnew is the length of the tube after it has been cut in half expressed in terms of the original length of the tube (L) therefore L(new) = L/2

(v) / (4(L/2))

Therefore f= v/2L

OK, it looks like you mean that L(new) = Lnew = L/2. So your reasoning seems to go like this:

When the length of the tube is $L$, the wavelength of the first harmonic is $L = \frac λ 2$. The original frequency of the first harmonic is then
$f_0 = \frac v λ = \frac v {2L}$

Cutting the tube in half reduces the frequency by the same factor, so
$L_{new} = \frac L 2 = \frac λ 4$

Then $f_0 = \frac v λ = \frac v {4 L_{new}}$

So now you have expressed the original frequency in terms of $L_{new}$. You have not said anything about what happens to the frequency when you reduce the length of the tube.

Now you substitute $\frac L 2$ for $L_{new}$ and get back your original equation
$f_0 = \frac v λ$

You still have not said anything about what happens to the frequency when you reduce the size of the tube, and you have not considered what happens when you close the end of the tube.

I suggest that you try this:
1) Consider what happens to the wavelength when you shorten the tube. Write an equation for the new wavelength $λ'$ in terms of $λ$.
2) Consider what happens to the wavelength when you close the end of the tube. Write an equation for this newer wavelength $λ''$ in terms of $λ'$.
3) Compare $λ''$ with $λ$.