A rope of length 2.00 m and mass 80 g is fixed at one end and tied...

The linear mass density of the rope is given by:

$\mu = \dfrac{m}{L} = \dfrac{80 \times 10^{-3}}{2.00} = 0.04 \text{ kg/m}$

The speed of the transverse wave on the rope is:

$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{256}{0.04}} = \sqrt{6400} = 80 \text{ m/s}$

The rope is fixed at one end, which acts as a node. The other end is tied to a light string (negligible mass), which provides no inertia and thus acts as a free end (antinode). The system behaves like a string fixed at one end and free at the other.

The allowed frequencies (harmonics) for such a system are odd multiples of the fundamental frequency, given by:

$f_n = \dfrac{(2n-1)v}{4L}$ for $n = 1, 2, 3, \dots$

The fundamental frequency ($n=1$) is:

$f_1 = \dfrac{v}{4L} = \dfrac{80}{4 \times 2.00} = 10 \text{ Hz}$

The first overtone corresponds to the third harmonic ($n=2$):

$f_2 = 3f_1 = 3 \times 10 = 30 \text{ Hz}$

The second overtone corresponds to the fifth harmonic ($n=3$):

$f_3 = 5f_1 = 5 \times 10 = 50 \text{ Hz}$

Thus, the frequencies are $10 \text{ Hz}$, $30 \text{ Hz}$, and $50 \text{ Hz}$.