Question

the speed of sound in air at 0°C is 332m/s. if it increases at the rate of 0.6m/s per degree, what will be the temperature when the velocity has increased to 334m/s?

Answer 1

Change in velocity is 12 m/s(v-u)
Rise after 1 degree increase is 0.6m/s
Therefore, rise after x degree increase is 12m/s
This is an example of direct variation
So, 1/x=0.6/12
X=12/0.6=120/6=20 degrees

Answer 2

The temperature when the velocity of the sound is 334 m/s is 3.3333°C

Given:

Speed of sound = 332 m/s

Rate of increase = 0.6 m/s per degree.

Solution:

So 1 degree rise in temperature from 0°C will increase the speed to $(332+0.6 (1))$ m/s

2 degree rise in temperature i.e T =2°C will increase the speed to $(332 + 0.6 (2))$ m/s.

In this way, we can derive a formula binding speed of sound with temperature as below

$\text{Speed of sound in air}T=\text { speed of sound in air at } 0^{\circ} \mathrm{C}+(\text {rate of increase perdegree } \times \text {Temperature})$

So according to the above equation,

In air, speed of sound at temperature T = 334 m/s

In air, sound speed at 0°C = 332 m/s

Rate of increase per degree = 0.6 m/s

Let, T be the temperature at which the speed increased to 334 m/s

Thus, T can be found by substituting the above values in the given equation

$334=332+0.6 T$

$334-332=0.6 T$

$2=0.6 T$

$\therefore T=\frac{2}{0.6}=3.3 \overline{3}$

Thus, In air, at 3.3333°C the sound speed reaches 334 m/s.