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Sound: Doppler Effect and Echo

11-MAR-2016 16:32

    Sound is a form of energy. It is that form of energy which makes us hear. Sound wave is a vibratory disturbance in a medium which carries energy from one point to another without there being a direct contact between the two points. Doppler Effect is an everyday experience. It is observed that the pitch of the sound is higher when we approach a stationary source of sound with high speed. And if we moved away from the source of sound the pitch becomes lower. This change in the pitch (frequency) of a wave due to motion of the source or observer is called Doppler Effect.

    Johann Christian Doppler, an Austrian physicist first proposed Doppler effect in 1842.   

     Jagranjosh

    The three situations under which frequency changes:

    Case 1: Source moving, observer stationary

    Let us choose the direction from the observer to the source as positive direction of velocity. Suppose the source (S) is moving with velocity (vs) and the observer and medium are at rest or stationary. Now, let the speed of wave at angular frequency and period (To), which is measured by the observer at rest in that medium be v. Let us assume that the observer has a detector which counts every time a wave crest reaches it.

    Now at time t = O the source is at point S1. The distance between source and observer at this point is L. At this point source emits a crest which reaches the observer at time t1 = L/v. Now at time t = To the source moves a distance vsTo and reaches at point S2. Th distance between observer and point S2 is L + vsTo. At point S2 the source emits second crest which reaches the observer at:

    t2 = To + (L + vsTo)/ v

    At time n To, the source emits (n+1)th crest and this reaches the observer at time:

    Tn+1   nTo    L ns To/v

    Hence in a time interval

    nTo      L nvs To/v    L/v

    the observer’s detector counts n crests and observer record period of the wave as T given by

    T      nTo     L  nvsTo/v    L/v/n

    =To+ vs To/v

    = To (1+vs/v)

    This equation can be rewritten in terms of the frequency vo:

    v = vo (1+vs/v)-1

    If vs is small compared with the wave speed v, then

    v = vo (1-vs/v)

    We replace vs with - vs, to get

    V = vo (1 + vs/v)

    Therefore, the observer notices higher frequency when the source approaches near him and lower frequency when the source moves away from him.

    Case 2: Observer moving; source stationary

    In this case observer is moving towards the source of sound and the source is at rest. Here the source and medium are approaching at speed vo and wave approaches with a speed of vo + v. Therefore the time interval between the arrival of the first and the (n+1)th crest is:

    tn+1 – t1 = n To – nvoTo/vo + v

    The observer measures the period of wave to be:

    = T (1- vo/ vo+v)

    T 1 vo/v -1

    Giving

    v = o(1+ vo/v)

    Case 3: Both source and observer moving

    Let the direction from observer to source be positive and source and observer be moving with velocities vs and vo. Let’s assume that at time t=O, observer is at point O1 and source is at point S1. When measured by the observer at rest in the given medium, the source emits wave of velocity v, of frequency v and period To. When source emits first crest then let the distance between O1 and S1 be L at t=O. Since the observer is moving the velocity of wave relative to observer is v+v. Therefore, the first crest reaches the observer at time t1= L (v+v).

    Now at time t = To, the observer and the source move to the position O2 and S2. The new distance between observer and source is L + (vs – vo) To. Source emits second crest at S2. This reaches the observer at time:

    t2 = To + (L + (vs – vo) To)/ (v + vo)

    The source emits its (n + 1)th crest at time nTo and this reaches the observer at time:

    tn+1 = nTo + (L + n(vs – vo) To)/ (v + vo)

    Hence, in a time interval tn+1 –t1, i.e.,

    nTo + (L+n(vs-vo)To) / (v+vo) – L / (v + vo),

    the crests counted by observer are n and period of wave recorded as equal to T, given by,

    T = To (1 + vs – vo/v + vo) = T (v + vs/v + vo)

    The frequency v observed by observer will be,

    V = vo (v + v/ v + vs)

    In case the observer and the source are moving with the same velocity then there will be no shift in frequency.

     Jagranjosh

    Uses of Doppler Effect

    • Used by police to check over speeding of vehicles.
    • At airport to guide aircraft.
    • In military to detect enemy aircraft.
    • Doctors use Doppler Effect to study heart beats and blood flow.

    Echo

    If stand in an empty hall and say something then after a while we will hear our own voice being reflected. We will hear our own sound in the form of echo.

    When the sound is repeated due to reflection of sound waves, it is called echo. Soft surface absorbs sound. So, it is when sound is reflected from a hard surface like tall brick wall or a cliff, we hear echo.

    Jagranjosh

    Calculation of minimum distance to hear an echo

    There has to be a time interval of 1/10th of a second or more for a human ear to hear two sounds separately. Therefore we can hear original sound and reflected sound only if there is a time-interval of at least 1/10th of a second or 0.1 second between them.

    Our distance from the sound reflecting surface should be 17.2 meters in order to hear an echo in air at a temperature of 20 degree. This distance will change as the temperature of air changes. Thus, echo will be more on hot day as compared to a cold day.

    The minimum distance to hear an echo in water should be 75 meters.

    Uses of echo

    • To measure depth of sea.
    • To locate underwater objects.
    • To investigate inside human body.

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