Question
What are the factors that affect the frequency of a vibrating string and how do they affect the frequency?
✓
Answer
The frequency (f) of a vibrating string is affected by:
(i) The length of the vibrating string. Frequency is inversely proportional to the length, i.e,$f \alpha \frac{1}{1}$. it
means, if length increases, frequency becomes less and if length decreases, frequency increases.
(ii) It is directly proportional to the square root of the tension, other factors remaining the same, e.g., if tension (or stretching force) is made 4 times, then the
frequency will be $(\sqrt{4}=2)$ doubled.
(iii) The frequency, (f) is inversely proportional to the square root of the mass of unit length of the vibrating
string, i.e., $f \alpha \frac{1}{\sqrt{ m ^{\prime}}}$ if $m ^{\prime}$ is the mass of unit length of
the string. Since $m^{\prime}=\pi r^2 . I$, it shows that $f$ is $\alpha \frac{1}{\sqrt{r^2}}$ or
$\frac{1}{ r }$, where $r$ is the radius of the vibrating strings.
(i) The length of the vibrating string. Frequency is inversely proportional to the length, i.e,$f \alpha \frac{1}{1}$. it
means, if length increases, frequency becomes less and if length decreases, frequency increases.
(ii) It is directly proportional to the square root of the tension, other factors remaining the same, e.g., if tension (or stretching force) is made 4 times, then the
frequency will be $(\sqrt{4}=2)$ doubled.
(iii) The frequency, (f) is inversely proportional to the square root of the mass of unit length of the vibrating
string, i.e., $f \alpha \frac{1}{\sqrt{ m ^{\prime}}}$ if $m ^{\prime}$ is the mass of unit length of
the string. Since $m^{\prime}=\pi r^2 . I$, it shows that $f$ is $\alpha \frac{1}{\sqrt{r^2}}$ or
$\frac{1}{ r }$, where $r$ is the radius of the vibrating strings.
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