[johnconnor92]

Can anyone please confirm with me my attempt at this question?thank you!

[XiaoXiao1]

If that is A squared then your final answer doesn't appear to be dimensionally correct; can you explain your fourth line?

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The acceleration I got = -2PAx/ml

[Stonebridge]

(Original post by johnconnor92)




Can anyone please confirm with me my attempt at this question?thank you!

You haven't explained where line 2 comes from.
You haven't explained where line 5 comes from.
To get line 2 you need to start from


expand the brackets and ignore the term with both delta p and delta V in it.

The force on the piston from the gas on the right is given by

F=δpA

Sub δp from your line 2 formula

sub δV = Aδl (δl is the displacement of the piston to the right)
Sub V=Al

Then acceleration = F/m

This will give you
a = -2pAδl / lm (there is a force from the gas on the left too)

This is SHM and you can compare with the simple pendulum to get the length of the equivalent pendulum.

[johnconnor92]

(Original post by Stonebridge)
You haven't explained where line 5 comes from.
To get line 2 you need to start from


expand the brackets and ignore the term with both delta p and delta V in it.

The force on the piston from the gas on the right is given by

F=δpA

Sub δp from your line 2 formula

sub δV = Aδl (δl is the displacement of the piston to the right)
Sub V=Al

Then acceleration = F/m

This will give you
a = -2pAδl / lm (there is a force from the gas on the left too)

This is SHM and you can compare with the simple pendulum to get the length of the equivalent pendulum.

Brilliant. Thank you so much Stonebridge. I owe you a big one.

[b_white]

Can I ask where you're getting these problems from? Thanks!