Question

When mass m is tied to the bottom of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 180 hz . adding an additional 1.2 kg to the hanging mass increases the second-harmonic frequency to 270 hz . part a what is m?

Answer 1

The frequency of the nth-harmonic of a string is given by
$f_n = \frac{n}{2L} \sqrt{ \frac{T}{\mu} }$
where n is the number of the harmonic, L is the length of the string, T the tension and $\mu$ the linear density. 

In our problem, since the mass m is tied to the string, the tension is equal to the weight of the object tied:
$T=mg$
Substituting into the first formula, we have
$f_n = \frac{n}{2L} \sqrt{ \frac{mg}{\mu} }$

In our problem we have n=2 (second harmonic). In the previous equation, the only factor which is not constant between the first and the second part of the problem is m, the mass. So, we can rewrite everything as
$f_2 = K \sqrt{m}$
where we called 
$K= \frac{2}{2L} \sqrt{ \frac{g}{\mu} }$

In the first part of the problem, the mass of the object is m and $f_2 = 180 Hz$. So we can write 
$180 Hz = K \sqrt{m}$

When the mass is increased with an additional 1.2 kg, the relationship becomes
$270 Hz = K \sqrt{(m+1.2 Kg)}$

By writing K in terms of m in the first equation, and subsituting into the second one, we get
$180 Hz \sqrt{ \frac{m+1.2 Kg}{m} }=270 Hz$
and solving this, we find
$m=0.96 kg$