A700.0-cm rope under a tension of 65.0 n is set into oscillation. the mass density of the rope is 90.0 g/cm. what is the frequency of the second overtone (n = 3)?

0.58 Hz . . . . (none of the offered choices)

The attachment shows the formula for the fundamental frequency. In order to use it, we need the various parameters to be in compatible units. For tension (T) in Newtons, the mass density (m/L) must be in kg/m.

We have a mass density of 90 g/cm, so ...

... (90 g/cm)×(1 kg/(1000 g))×(100 cm/m) = 9 kg/m

Likewise, the length (L) must be in meters:

... (700 cm)×(1 m/(100 cm)) = 7 m

Then the fundamental frequency (f₁) is ...

... f₁ = √(T/(m/L))/(2L) = √(65/9)/(2·7) = 0.192 . . . Hz

The 2nd overtone (3rd harmonic) is 3 times this frequency:

... f₃ = 3·f₁ ≈ 3·0.192 Hz = 0.576 Hz

_____

Comment on answer choices

Very likely, your computer is expecting you to select 0.18 Hz. This number is the number you get if you use the mass density of 90 g/cm directly, without changing it to kg/m first. That is, the correct answer (0.58 Hz) is inappropriately divided by √10 to get 0.18 "Hz." (The units are not actually Hz, but are Hz×√10.)

Have your teacher show you the dimensional analysis that goes along with arriving at the number 0.18 "Hz".

You can't do this problem the way the units are now, and when you do try it as it stands, you can't get the answers they are getting. Since you are likely doing a review, you might be intended to get the a different answer than the ones listed so that you can make an educated appeal. You can proceed in one of two ways but you must choose one of them.

The rope is in gram / cm so you could change the 65 N into dynes  

Formula

f = sqrt(T / (m/L )) * 2 * L

Solution

The other way to do it to convert the 90 grams / cm into kg / m

90 grams / cm * [ 1 kg/1000 grams] * [100 cm / 1 m]

90 grams / cm = 9 kg / m

Givens

Solution

f3 = 0.576 Hz In any case the answer is not there and I'm reasonably certain that two of us doing the same question getting more or less the same answer cannot be totally incorrect unless we are both making the same mistake.