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3 months ago

Waves Class 11th

An observer is riding on a bicycle and moving towards a hill at 18 kmh^-1. He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is 640 Hz and velocity of the sound in air is 320 m/s, the beat frequency between the two sounds heard by observer will be _________ Hz.

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Vishal Baghel

Contributor-Level 10

$v_{S} = 0, v_{O_{b}} = 5 m / s$

$f_{d i r e c t} = (\frac{320 - 5}{320}) 640 = 630 H z$

$f_{b e a t} = (650 - 630) = 20 H z$

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3 months ago

Waves Class 11th

The frequency of echo will be_________Hz if the train bellowing a whistle of frequency 320 Hz is moving with a velocity of 36 km/h towards a hill from which an echo is heard by the train driver. Velocity of sound in air is 330 m/s.

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Vishal Baghel

Contributor-Level 10

$υ_{0} = 320 H z . v_{t r a i n} = 36 k m / h = \frac{36 * 1000}{60 * 60} = 10 m / s$

$υ_{(r e f l e c t e d S o u n d)} = υ_{0} (\frac{v}{v - v_{t r a i n}})$

$= 320 (\frac{330}{330 - 10}) = 330 H z$

$\Rightarrow υ_{e c h o} = υ_{r e f l e c t e d} (\frac{v + v_{t r a i n}}{V})$

$= 330 (\frac{330 + 10}{330}) = 340 H z .$

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3 months ago

Waves Class 11th

The equations of two waves are given by :

$\begin{array}{l} y_{1} = 5 s i n 2 π (x - υ t) c m \\ y_{2} = 3 s i n 2 π (x - υ t + 1.5) c m \end{array}$

These waves are simultaneously passing through a string. The amplitude of the resulting wave is :

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Vishal Baghel

Contributor-Level 10

$y_{1} = 5 s i n 2 π (x - v t) c m$

$y_{2} = 3 s i n 2 π (x - v t + 1.5) c m$

$? = (p h a s e A n g l e) = 2 π * 1.5 = 3 π$

$y_{R} = \sqrt[]{y_{1}^{2} + y_{2}^{2} + 2 y_{1} y_{2} c o s ?}$

$\begin{array}{l} = \sqrt[]{5^{2} T 3^{2} + 2 * 5 * 3 c o s 3 π} \\ = \sqrt[]{25 + 9 + (30 * - 1)} \end{array}$

$\begin{array}{l} = \sqrt[]{34 - 30} \\ = \sqrt[]{4} \\ = 2 c m \end{array}$

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3 months ago

Waves Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

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:

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alok kumar singh

Contributor-Level 10

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?

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3 months ago

Waves Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

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:

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alok kumar singh

Contributor-Level 10

Maximum particle velocity = $A ω$

Wave velocity = $\frac{ω}{R}$

$\Rightarrow \frac{ω}{k} = A ω$

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

$λ = \frac{2 π}{k} = 4 π c m$

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4 months ago

Waves Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

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:

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alok kumar singh

Contributor-Level 10

(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, $m = 1, n = 5, λ = 1$

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

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4 months ago

Waves Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

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

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Payal Gupta

Contributor-Level 10

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

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

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

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4 months ago

Waves Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

The driver of bus approaching a big wall notices that the frequency of his bus's horn changes from $420 H z$ to $490 H z$ when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is $330 {m s}^{- 1}$ :

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New answer posted

4 months ago

Waves Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Which of the following statements are true for a stationary wave?

(a) Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.

(b) All the particles cross their mean position at the same time.

(c) All the particles are oscillating with same amplitude.

(d) There is no net transfer of energy across any plane.

(e) There are some particles which are always at rest.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, d, e) a) clearly every particle at x will have amplitude =asinkx=fixed

b) for mean position =0

coswt=0

wt= (2n-1) $\frac{π}{2}$

hence for a fixed of n all particles are having same value of time t= (2n-1) $\frac{π}{2 w}$

c) amplitude of all the particles are asinkx which is different for different particles at different values of x

d) the energy is a stationary wave is confined between two nodes.

e) particles at different nodes are always at rest.

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4 months ago

Waves Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10m/s. Given that the speed of sound in still air is 340m/s,

(a) The frequency of sound as heard by an observer standing on the platform is 400Hz.

(b) The speed of sound for the observer standing on the platform is 350m/s.

(c) The frequency of sound as heard by the observer standing on the platform will increase.

(d) The frequency of sound as heard by the observer standing on the platform will decrease.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) v_o=400Hz, v=340m/s

V_m=10m/s

(a) as both source and observer are stationary, hence frequency observed will be same as natural frequency v_o=400Hz

(b) the speed of sound v=v+v_w= 340+10=350m/s

(c) there will be on effect on frequency because there is no relative motion between source and observer hence c, d are incorrect.

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