Question 455: Waves

Question

Two identical strings of length $$l$$ and $$2l$$ vibrate with fundamental frequencies $$\mathrm{N} \mathrm{~Hz}$$ and $$1.5 N$$ Hz, respectively. The ratio of tensions for smaller length to large length is

Accepted Answer

The fundamental frequency of a vibrating string can be expressed as:
$$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$
where:
$f$ is the frequency of vibration,
$L$ is the length of the string,
$T$ is the tension in the string, and
$\mu$ is the linear mass density of the string (mass per unit length).
From the given question, we have two strings with lengths $l$ and $2l$ vibrating at fundamental frequencies $N$ Hz and $1.5N$ Hz, respectively. Let's denote the tension in the shorter string as $T_1$ and in the longer string as $T_2$. Since the strings are identical, their linear mass densities ($\mu$) are the same.
The frequency of the shorter string ($l$) is given as:
$$N = \frac{1}{2l} \sqrt{\frac{T_1}{\mu}}$$
The frequency of the longer string ($2l$) is given as:
$$1.5 N = \frac{1}{2(2l)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4l} \sqrt{\frac{T_2}{\mu}}$$
We can manipulate these equations to solve for the ratio of tensions $T_1$ and $T_2$. First, rearrange the formula of the frequency for $T_1$ and $T_2$:
For the small string:
$$T_1 = \mu N^2 (2l)^2 = 4l^2\mu N^2$$
For the large string:
$$T_2 = \mu (1.5N)^2 (4l)^2 =16l^2 \mu (1.5N)^2 = 16l^2 \mu (2.25N^2) = 36l^2\mu N^2$$
The ratio of tensions $T_1 : T_2$ can be found as:
$$\frac{T_1}{T_2} = \frac{4l^2\mu N^2}{36l^2\mu N^2} = \frac{4}{36} = \frac{1}{9}$$
Therefore, the ratio of tensions for smaller length to larger length is $1:9$, which corresponds to
Option B
.

Answer

$$1: 3$$

Answer

$$1: 9$$

Answer

$$3: 1$$

Answer

$$9: 1$$

Waves

Two identical strings of length $l and 2l vibrate with fundamental frequencies \mathrm{N} \mathrm{~Hz} and 1.5 N$ Hz, respectively. The ratio of tensions for smaller length to large length is

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