Question 87: Waves

Question

In an experiment with sonometer when a mass of $$180 \mathrm{~g}$$ is attached to the string, it vibrates with fundamental frequency of $$30 \mathrm{~Hz}$$. When a mass $$\mathrm{m}$$ is attached, the string vibrates with fundamental frequency of $$50 \mathrm{~Hz}$$. The value of $$\mathrm{m}$$ is ___________ g.

Accepted Answer

We can use the fact that the ratio of frequencies is equal to the square root of the ratio of tensions:
$$\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}$$
In the first case, the mass attached to the string is $$180 \mathrm{~g}$$ and the frequency is $$30 \mathrm{~Hz}$$, so we have:
$$\frac{f_2}{30~\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$$
In the second case, the frequency is $$50 \mathrm{~Hz}$$, so we have:
$$\frac{50~\mathrm{Hz}}{30~\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$$
Simplifying, we get:
$$\frac{5}{3}=\sqrt{\frac{T_2}{T_1}}$$
Squaring both sides, we get:
$$\frac{25}{9}=\frac{T_2}{T_1}$$
Since the tension in the string is proportional to the mass attached to it, we can write:
$$\frac{m}{180~\mathrm{g}}=\frac{T_2}{T_1}=\frac{25}{9}$$
Solving for $$m$$, we get:
$$m=\frac{25}{9}(180~\mathrm{g})=\boxed{500~\mathrm{g}}$$
Therefore, the mass attached to the string in the second case is $$500 \mathrm{~g}$$.

Waves

In an experiment with sonometer when a mass of $180 \mathrm{~g} is attached to the string, it vibrates with fundamental frequency of 30 \mathrm{~Hz}. When a mass \mathrm{m} is attached, the string vibrates with fundamental frequency of 50 \mathrm{~Hz}. The value of \mathrm{m}$ is ___________ g.

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