**frequency**and

**tension**in the string increase.

Â

There are two formulae to calculate the waves on the string and these are:

\[Â v = \lambda fÂ \]

\[Â v =Â \sqrt { \frac { T } { \mu }} \]

Here, **v** is the **speed** of the wave in the string, **f** represents the **frequency** of that wave, **T** is the **tension** produced in the string, and $ \mu $ represents the mass per unit length of the string. Considering a standard straight string having the **mass and length** both **constant**, we have to find the tension and frequency of that string.

## Expert Answer

We can **increase** the tension in the string if we put the **frequency constant** in **case 1** and we can calculate the effect of this **increase in tension** on the other variables used in the formulae like $ \lambda $, $ v $, $ f $, $ T $ and $ \mu $

**Two weights** are used to calculate the **increase in tension** of the spring. Two weights are suspended to the hook attached to the spring. The following effect on the variables occurred:

\[ v \propto T \]

According to the given expression of **velocity** and tension, the velocity is **directly proportiona**l to the tension in the string. If the velocity increases, the tension in the spring also increases.

$ \lambda $ represents the **wavelength** which is **directly proportional** to the tension in the string. The increase in one quantity causes an increase in another quantity.

\[ \mu = constant \]

**Mass** per unit length of the string will be **constant** as given in the question.

\[ f = constant \]

The frequency of the waves in the string will be constant as given.

The **frequency of the waves** in the string can be increased by changing the i**nput frequency** on the **frequency generator** and studying the effect of this frequency on the other variables used in the formulae like $ \lambda $, $ v $, $ f $, $ T $ and $ \mu $.

By changing the frequency:

\[ vÂ \propto f \]

**Velocity increases** as the frequency increases because velocity is directly proportional to the frequency of the waves.

\[ f \propto \frac { 1 } { \lambda } \]

$ \lambda $ decreases with the increase in the frequency of the wave as it is** inversely proportional** to the frequency.

\[ \mu = constant \]

Mass per unit length of the string will be constant with the increase of the frequency as given in the question.

\[ T = constant \]

The tension in the string will be constant as given in the question.

## Numerical Results

**The increase in the tension causes an increase in wavelength and velocity while the increase in frequency causes a decrease in wavelength and an increase in velocity.**

## Example

Study the effect on the string if $ \lambda $ increases by keeping the frequency constant.

By changing the frequency:

\[ vÂ \propto \lambda \]

Velocity increases as the wavelength increases because velocity is **directly proportional** to the wavelength of the waves.

\[ \lambda \propto \frac { 1 } { f } \]

$ \lambda $ increases with the decrease in the frequency of the wave as it is inversely proportional to the frequency.

\[ \mu = constant \]

Mass per unit length of the string will be constant with the **increase of the frequency** as given in the question.

\[ T = constant \]

The **tension** in the string will be **constant** as given in the question.