# The wave speed on a string under tension is 200 m/s. What is the speed if he tension is doubled?

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:

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

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:

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

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

## 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:

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

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