The **aim of this question** is to understand the key concepts of **speed, frequency, wavelength, and tension in a string.**

Whenever** energy is transferred** from one place to another through the **successive vibratory motion of particles**, this form of energy transferring agent is **called a wave. **All type of waves have some common properties such as the **speed, frequency, wavelength etc.**

The** speed of a wave traveling through a string** depends upon its **tension **$ F_{ T } $, **mass of the string** $ m $, and the **length of the string** $ L $. Given these parameters, it can be** calculated using the following formula**:

\[ v_{ wave } \ = \ \sqrt{ \dfrac{ F_{ T } \times L }{ m } } \]

## Expert Answer:

**Lets say:**

\[ \text{ speed of wave at original tension } \ = \ v_{ wave } \ = \ \sqrt{ \dfrac{ F_{ T } \times L }{ m } } \]

\[ \text{ speed of wave at doubled tension } \ = \ v’_{ wave } \ = \ \sqrt{ \dfrac{ 2 \times F_{ T } \times L }{ m } } \]

Notice that both $ L $ and $ m $ **remain the same** because they are the **property of the string,** which is not changed. **Dividing both of the above equations:**

\[ \dfrac{ v’_{ wave } }{ v_{ wave } } \ = \ \dfrac{ \sqrt{ \dfrac{ 2 \times F_{ T } \times L }{ m } } }{ \sqrt{ \dfrac{ F_{ T } \times L }{ m } } } \]

\[ \Rightarrow \dfrac{ v’_{ wave } }{ v_{ wave } } \ = \ \sqrt{ \dfrac{ 2 \times F_{ T } \times L \times m }{ F_{ T } \times L \times m } } \]

\[ \Rightarrow \dfrac{ v’_{ wave } }{ v_{ wave } } \ = \ \sqrt{ 2 } \]

\[ \Rightarrow v’_{ wave } \ = \ \sqrt{ 2 } v_{ wave } \ … \ … \ … \ … \ ( 1 ) \]

**Substituting values:**

\[ \Rightarrow v’_{ wave } \ = \ \sqrt{ 2 } ( 200 \ m/s ) \]

\[ \Rightarrow v’_{ wave } \ = \ 280 \ m/s \]

Which is the **required answer**.

## Numerical Result

\[ \Rightarrow v’_{ wave } \ = \ 280 \ m/s \]

## Example

What happens to the** speed of the wave** if the **tension in the string is raised by four times** instead of doubling?

**Lets say:**

\[ \text{ speed of wave at original tension } \ = \ v_{ wave } \ = \ \sqrt{ \dfrac{ F_{ T } \times L }{ m } } \]

\[ \text{ speed of wave at four times the tension } \ = \ v’_{ wave } \ = \ \sqrt{ \dfrac{ 4 \times F_{ T } \times L }{ m } } \]

**Dividing both of the above equations:**

\[ \dfrac{ v’_{ wave } }{ v_{ wave } } \ = \ \dfrac{ \sqrt{ \dfrac{ 4 \times F_{ T } \times L }{ m } } }{ \sqrt{ \dfrac{ F_{ T } \times L }{ m } } } \]

\[ \Rightarrow \dfrac{ v’_{ wave } }{ v_{ wave } } \ = \ \sqrt{ \dfrac{ 4 \times F_{ T } \times L \times m }{ F_{ T } \times L \times m } } \]

\[ \Rightarrow \dfrac{ v’_{ wave } }{ v_{ wave } } \ = \ \sqrt{ 4 } \]

\[ \Rightarrow \dfrac{ v’_{ wave } }{ v_{ wave } } \ = \ 2 \]

\[ \Rightarrow v’_{ wave } \ = \ 2 v_{ wave } \ … \ … \ … \ … \ ( 2 ) \]

**Substituting values:**

\[ \Rightarrow v’_{ wave } \ = \ 2 ( 200 \ m/s ) \]

\[ \Rightarrow v’_{ wave } \ = \ 400 \ m/s \]