Calculate L using Omega, Lo = 16, and h = 1.5

[eddievic]

Find L if...

Homework Statement

$$\omega=\frac{1}{h}In(\frac{L}{Lo}-1)$$

Find L if omega - -2.6, Lo = 16 and h = 1.5

Homework Equations

The Attempt at a Solution

-2.6*1.5=In(L/16-1)
-3.9=In(L/16-1
e^-3.9=L/16-1
0.02024=L/16-1
1.02024=L/16
so L=16.32384

 

[SteamKing]

The correct abbreviation for the natural log is 'lower case L' and 'lower case N'. You have 'Upper case I' and 'lower case N', which is incorrect. Otherwise, your algebra appears to be correct.

 

[eddievic]

The correct abbreviation for the natural log is 'lower case L' and 'lower case N'. You have 'Upper case I' and 'lower case N', which is incorrect. Otherwise, your algebra appears to be correct.

Ok thanks for the help I will use the correct abbreviation in future

 

[Mark44]

Homework Statement

$$\omega=\frac{1}{h}In(\frac{L}{Lo}-1)$$

Find L if omega - -2.6, Lo = 16 and h = 1.5

Homework Equations

The Attempt at a Solution

-2.6*1.5=In(L/16-1)
-3.9=In(L/16-1
e^-3.9=L/16-1
0.02024=L/16-1
1.02024=L/16
so L=16.32384

It's best to substitute the constants at the end of your algebraic manipulation. Doing so sooner can lead to round-off errors.

This is the way I would have done it.
$$\omega=\frac{1}{h}\ln(\frac{L}{L_0}-1)$$
$$\Rightarrow h \omega = \ln(\frac{L}{L_0}-1)$$
$$\Rightarrow e^{h \omega} = \frac{L}{L_0}-1$$
$$\Rightarrow e^{h \omega} + 1 = \frac{L}{L_0}$$
$$\Rightarrow L_0 (e^{h \omega} + 1) = L$$
Now you can replace the constants.

 

[eddievic]

It's best to substitute the constants at the end of your algebraic manipulation. Doing so sooner can lead to round-off errors.

This is the way I would have done it.
$$\omega=\frac{1}{h}\ln(\frac{L}{L_0}-1)$$
$$\Rightarrow h \omega = \ln(\frac{L}{L_0}-1)$$
$$\Rightarrow e^{h \omega} = \frac{L}{L_0}-1$$
$$\Rightarrow e^{h \omega} + 1 = \frac{L}{L_0}$$
$$\Rightarrow L_0 (e^{h \omega} + 1) = L$$
Now you can replace the constants.

this makes a lot of sense thanks Mark44