Question

State two ways by which the frequency of the transverse vibrations of a stretched string can be increased.

Answer 1

Hint : This question can be solved by determining the factors on which the frequency of transverse vibrations of a stretched string depends on. The factors are altered accordingly in order to increase the frequency.

Formula Used : The following formulae have been used,
 $ \Rightarrow {\text{f = }}\dfrac{{\text{1}}}{{{\text{2l}}}}\sqrt {\dfrac{{\text{T}}}{{\text{m}}}} $ where $ {\text{f}} $ is the frequency, $ {\text{l}} $ is the length of the vibrating string, $ {\text{T}} $ is the tension in the string and $ {\text{m}} $ is the mass per unit length of the string.
 $\Rightarrow f = \dfrac{{\text{1}}}{{{\text{2l}}}}\sqrt {\dfrac{{\text{T}}}{{{{\pi }}{{\text{r}}^{\text{2}}}{\text{d}}}}} $ where $ {\text{f}} $ is the frequency, $ {\text{l}} $ is the length of the vibrating string, $ {\text{T}} $ is the tension in the string, $ {\text{r}} $ is the radius and $ {\text{d}} $ is the density of the string.

Complete step by step answer
The number of vibrations produced by a particle in one second is known as frequency. Each string has a specific value known as natural frequency. The factors that affect the frequency of the string is its length, its diameter, tension and its density.
The frequency of a transverse vibration is given by the formula, $ {\text{f = }}\dfrac{{\text{1}}}{{{\text{2l}}}}\sqrt {\dfrac{{\text{T}}}{{\text{m}}}} $ where $ {\text{f}} $ is the frequency, $ {\text{l}} $ is the length of the vibrating string, $ {\text{T}} $ is the tension in the string, $ {\text{m}} $ is the mass per unit length of the string.
It can also be written as, $ {\text{f = }}\dfrac{{\text{1}}}{{{\text{2l}}}}\sqrt {\dfrac{{\text{T}}}{{\text{m}}}} = \dfrac{{\text{1}}}{{{\text{2l}}}}\sqrt {\dfrac{{\text{T}}}{{{{\pi }}{{\text{r}}^{\text{2}}}{\text{d}}}}} $
Since, mass ( $ {\text{m}} $ ) is the product of volume ( $ {\text{V}} $ ) and density ( $ {\text{d}} $ ),
 $ \Rightarrow {\text{m = V}} \times {{d = \pi }}{{\text{r}}^{\text{2}}}{\text{d}} $ , where $ {\text{r}} $ is the radius and $ {\text{d}} $ is the density of the string.
The frequency of transverse vibrations can be increased in the following ways:
The frequency can be increased by decreasing the length of the string. This is because $ {\text{f}} \propto \dfrac{{\text{1}}}{{{\text{2l}}}} $ , where f is the frequency and l is the length of the string. The length of string affects frequency in a way that shorter lengths have high frequency and longer lengths have lower frequencies.
The frequency of transverse vibrations of a string can also be increased by decreasing the diameter of a string. The diameter of a string is its thickness- the diameter is equal to the twice the radius of the string. The radius of the string $ {\text{r}} $ is inversely proportional to the frequency of the string, $ {\text{f}} \propto \sqrt {\dfrac{{\text{1}}}{{{{\text{r}}^{\text{2}}}}}} $ . Hence, when a string has a larger diameter it vibrates with a lower frequency and for a smaller diameter of the string, the frequency is higher.

Note
The frequency of a string also depends on other factors as tension on the string and its density.
Tension is the force exerted by a string, along its length; it is an action-reaction pair of forces acting at the ends of a string. The greater the tension in the string, the more tightly it is stretched.
As, $ {\text{f}} \propto \sqrt {\text{T}} $ where f is frequency and T is tension, thus for greater tension, frequency will be higher and for lower tension of the string, we have lower frequencies.
The density of a string affects frequency in a way such that denser strings vibrate at slower speeds, thus low frequencies. Strings that are less dense vibrate faster, with higher frequencies. The frequency is inversely proportional to the square root of the density of the string, $ {\text{f}} \propto \sqrt {\dfrac{1}{d}} $ where $ {\text{f}} $ is the frequency and $ {\text{d}} $ is the density of the string.