Question: In Melde's experiment, when wire is stretched by empty pan, four loops are obtained and when six gra...

Question

In Melde's experiment, when wire is stretched by empty pan, four loops are obtained and when six gram weight is added in the pan, the number of loops becomes one. The mass of pan is

Answer 1

Solution

Let the mass of the pan be $m$ grams. In Melde’s experiment with fixed frequency $f$ and string length $L$ , the number of loops $n$ is inversely proportional to the wavelength and can be shown to vary inversely with the square root of tension $T$ . Since the tension is supplied by the weights, we have:

n \propto \frac{1}{\sqrt{T}}

For the two cases:

Empty pan:
Tension $T_1 = m\,g$ and number of loops $n_1 = 4$ .
Pan with 6 g weight:
Tension $T_2 = (m + 6)\,g$ and number of loops $n_2 = 1$ .

Setting up the proportionality ratio:

\frac{n_1}{n_2} = \sqrt{\frac{T_2}{T_1}} \quad \Longrightarrow \quad \frac{4}{1} = \sqrt{\frac{m+6}{m}}

Squaring both sides gives:

16 = \frac{m+6}{m}

Solving for $m$ :

16m = m + 6 \quad \Longrightarrow \quad 15m = 6 \quad \Longrightarrow \quad m = \frac{6}{15} = 0.4 \text{ grams}