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Physics Ncert Solutions Class 12th Physics Class 12th Waves

A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency (in Hz) of the source when it is moving away from the observer after crossing him? (Take speed of sound in air as 350 m/s)

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Vishal Baghel | Contributor-Level 10

3 weeks ago

Kindly consider the following figure

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In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning form of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms^-1. The third resonance is observed when the air column has a length of ____________cm.

Vishal Baghel

According to Concept of resonance tube, we can write

$\frac{λ}{4} + e = l_{1} a n d \frac{3 λ}{4} + e = l_{2} \Rightarrow \frac{λ}{2} = l_{2} - l_{1} \Rightarrow \frac{V}{2 ν} = l_{2} - l_{1}$

$\Rightarrow l_{2} = \frac{v}{2 ν} + l_{1} = \frac{336}{400} + 0.20 = 0.84 + 0.20 = 1.04 m = 104 c m$

A longitudinal wave is represented by x = 10 sin 2p $(n t - \frac{x}{λ}) c m .$ The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

alok kumar singh

x = 10 $2 ? (n t ? \frac{x}{?}) c m$

V (wave velocity) = $\frac{? ?}{2 ?}$

v_max = 10 * 2p n

10 * 2pn = 4 * $\frac{? ?}{2 ?}$

10 * 2pn = $4 \frac{2 ? ? n ?}{2 ?}$

x = 5?

A transverse wave s represented by y = $2 s i n (ω t - k x) c m .$ The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity. Will be:

alok kumar singh

Maximum particle velocity = $A ω$

Wave velocity = $\frac{ω}{R}$

$\Rightarrow \frac{ω}{k} = A ω$

$\Rightarrow k = \frac{1}{A} = \frac{1}{2} = c m$

$λ = \frac{2 π}{k} = 4 π c m$

For a transverse wave traveling along a straight line, the distance between two peaks (crests) is 5
, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in $A \to I V; B \to I$ ) of the waves are:

alok kumar singh

(n) $λ = 5$

$(n, m)$ : Integers

$\begin{array}{r} (\frac{2 m + 1}{2}) λ = \frac{3}{2} \\ \Rightarrow \frac{3 / 2}{5} = \frac{2 m + 1}{2 n} \\ \Rightarrow 3 n = 10 m + 5 \end{array}$

$N, m$ are integers.

So, $m = 1, n = 5, λ = 1$

$\begin{array}{l} m = 4 & n = 15 & λ = \frac{1}{3} \\ m = 7 & n = 25 & λ = \frac{1}{5} \end{array}$

The velocity of sound in a gas, in which two wavelengths 4.08m and 4.16m produce 40 beats in 12s, will be

Payal Gupta

$| f_{1} - f_{2} | = \frac{40}{12} = \frac{10}{3}$

$\Rightarrow v (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = \frac{10}{3}$

$\Rightarrow$ $V$ $=$ $\frac{10}{3}$ $\times$ $\frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}}$ $=$ $\frac{10}{3}$ $\times$ $\frac{4.08 \times 4.16}{. 08}$ $=$ $\frac{10}{3}$ $\times$ $\frac{408 \times 4.16}{8}$ $=$ $7$ $0$ $7$ $.$ $2$ $m$ $/$

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