Answer

Verified

420k+ views

**Hint:**When an observer moves towards or away from the source of sound, different frequencies are observed due to Doppler Effect. When the source moves towards the observer, the frequency increases and when the source moves away from the observer, the frequency decreases.

**Complete step by step answer:**

Doppler effect is observed in everyday life, when standing on a station; the approaching train’s siren feels to be having greater frequency. When the same trains pass us, the frequency seems to be decreasing.

The formula to calculate the apparent frequency is:

$$f = \left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right){f_0}$$

Here, $${f_0}$$ is the actual frequency that the source produces.

$$f$$ is the apparent frequency observed by the observer;

$$v$$ is the velocity of sound;

$${v_o}$$ is the velocity of the observer;

$${v_s}$$ is the velocity of the source.

The given values are:

$${v_o} = 0$$ as the observer is at rest;

$$v = 340\,m\,{s^{ - 1}}$$ given speed of sound in air;

$${f_0} = 60\,Hz$$ frequency at which the whistle is blown;

$${v_s} = 72\,km\,h{r^{ - 1}}\, = \,20\,m\,{s^1}$$ speed at which the train is moving towards the observer.

As the source is moving towards the observed, hence greater frequency is observed. Therefore, the above formula is modified as:

$$f = \left( {\dfrac{{v \pm {v_o}}}{{v - {v_s}}}} \right){f_0}$$

A decrease in denominator will increase the resultant frequency.

Substituting these values in above equation, we have:

$$f = \left( {\dfrac{{340 \pm 0}}{{340 - 20}}} \right)640$$

$$ \Rightarrow f = \dfrac{{340}}{{320}} \times 640$$

$$ \Rightarrow f = 680\,Hz$$

This is the frequency when the train is moving towards the observer.

When the train is moving away from the observer, the apparent frequency must decrease. Therefore, the formula is modified as:

$$f = \left( {\dfrac{{v \pm {v_o}}}{{v + {v_s}}}} \right){f_0}$$

As the denominator will increase, the overall value will decrease.

Substituting the values, we have:

$$f = \left( {\dfrac{{340 \pm 0}}{{340 + 20}}} \right) \times 640$$

$$ \Rightarrow f = \left( {\dfrac{{340}}{{360}}} \right) \times 640$$

$$ \Rightarrow f = 604.4\,Hz$$

This is the frequency when the train is moving away from the observer.

The frequency when the train moves towards and away from the observer is $$680\,Hz$$ and $$604.4\,Hz$$ respectively.

**Note:**

The Doppler effect is observed as the source starts moving towards the observer, the distance between them decreases and hence the sound wave reaches faster. While solving, convert the speed of the train in SI units. Also remember that if the observer would have been moving then we would have modified the equation accordingly.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference Between Plant Cell and Animal Cell

Which are the Top 10 Largest Countries of the World?

10 examples of evaporation in daily life with explanations

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE