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A wave travelling in the positive x-direction having maximum displacement along y-direction as 1m, wavelength 2π m and frequency of 1/π Hz is represented by

y=sin(x-2t)

y=sin(2πx-2πt)

y=sin(10πx-20πt)

y=sin(2πx+2πt)

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Explanation

(a) Given a=1m

As y=a sin(kx-ωt)

=sin(2π/2π x-2π x 1/π t)

=sin (x-2t)

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If we study the vibration of a pipe open at both ends. then the following statements is not true

Open end will be anti-node

Odd harmonics of the fundamental frequency will be generated

All harmonics of the fundamental frequency will be generated

Pressure change will be maximum at both ends

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A source of unknown frequency gives 4 beats/s when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency 513 Hz. The unknown frequency is

254 Hz

246 Hz

240 Hz

260 Hz

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When a string is divided into three segments of lengths $l_{1}, l_{2} a n d l_{3},$ the fundamental frequencies of these three segments are $v_{1}, v_{2} a n d v_{3}$ respectively. The original fundamental frequency (v) of the string is

$\sqrt{v} = \sqrt{v_{1}} + \sqrt{v_{2}} + \sqrt{v_{3}}$

$v = v_{1} + v_{2} + v_{3}$

$\frac{1}{v} = \frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}$

$\frac{1}{\sqrt{v}} = \frac{1}{\sqrt{v_{1}}} + \frac{1}{\sqrt{v_{2}}} + \frac{1}{\sqrt{v_{3}}}$

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Explanation

The fundamental frequency of string

$v = \frac{1}{2 l} \sqrt{\frac{T}{m}}$

$∴ v_{1} l_{1} = v_{2} l_{2} = v_{2} l_{3} = k$ ...(i)

From Eq. (i)

$l_{1} = \frac{k}{v_{1}}, l_{2} = \frac{k}{v_{2}}, l_{3} = \frac{k}{v_{3}}$

Original length

$l = \frac{k}{v}$

Here, $l = l_{1} + l_{2} + l_{3}$

$\frac{k}{v} = \frac{k}{v_{1}} + \frac{k}{v_{2}} + \frac{k}{v_{3}}$

$\frac{1}{v} = \frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{k}{v_{3}}$

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Two sources of sound placed close to each other, are emitting progressive waves given by

$y_{1}$ =4 sin 600 $πt$ and $y_{2}$ =5 sin 608 $πt$

An observer located near these two sources of sound will hear

4 beats per second with intensity ratio 25:16 between waxing and waning

8 beats per second with intensity ratio 25:16 between waxing and waning

8 beats per second with intensity ratio 81:1 between waxing and waning

4 beats per second with intensity ratio 81:1 waxing and waning

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Explanation

Given, $y_{1} = 4 \sin 600 πt$

and $y_{2} = 5 \sin 608 πt$

Comparing with general equation

$y = a \sin 2 πft$

we get, $f_{1} = 300 Hz and f_{2} = 304 Hz$

Number of beats = $f_{2} - f_{1} = 4 s^{- 1}$

$\frac{I_{\max}}{I_{\min}} = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = {(\frac{4 + 5}{4 - 5})}^{2} = \frac{81}{1}$

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The equation of a simple harmonic wave is

given by

$y = 3 \sin \frac{π}{2} (50 t - x)$

where x and y are in meters and t is in

seconds. The ratio of maximum particle

velocity to the wave velocity is

$2 π$

$\frac{3}{2} π$

$3 π$

$\frac{2}{3} π$

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Explanation

We know that

$v_{\max} = aω and v = nλ so, \frac{v_{\max}}{v} = \frac{aω}{nλ} = \frac{a (2 πn)}{nπ} = \frac{2 πa}{λ} = \frac{2 πa}{2 π / k} = ka = \frac{π}{2} \times 3 = \frac{3 π}{2}$

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A train moving at a speed of 220 ${ms}^{- 1}$

towards a stationary object, emits a sound

of frequency 1000 Hz. Some of the sound

reaching the object gets reflected back to

the train as echo. The frequency of the echo

as detected by the driver of the train is

(speed of sound in air is 330 ${ms}^{- 1}$ )

3500Hz

4000Hz

5000Hz

3000Hz

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Explanation

From Doppler's shift, we know for this case

$Frequency recieved by stationary observer be n_{1} n_{1} = n (\frac{v}{v - v_{s}}) Apparent frequency for driver n^{'} = n_{1} (\frac{v + v_{s}}{v}) n' = n (\frac{v + v_{s}}{v - v_{s}}) = 1000 (\frac{330 + 220}{330 - 220}) = 1000 (\frac{550}{110}) = 5000 Hz$

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Two waves are represented by the equations

$y_{1} = a \sin (ωt + kx + 0.57) m$ and

$y_{2} = a \cos (ωt + kx) m$ , where x is in metre

and t in second. The phase difference between

them is

1.25 rad

1.57 rad

0.57 rad

1.0 rad

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Explanation

$y_{1} = a \sin (ωt + kx + 0.57) m$

and $y_{2} = a \cos (ωt + kx) m$

or $y_{2} = a \sin (\frac{π}{2} + ωt + kx) m$

Phase difference

$∆ ϕ = ϕ_{2} - ϕ_{1} = \frac{π}{2} - 0.57 = 1.57 - 0.57 = 1 rad$

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Sound waves travel at 350 m/s through a warm

air and at 3500 m/s through brass. The wavelength

of a 700 Hz acoustic wave as it enters brass from

warm air :

increases by factor 20

increases by factor 10

decreases by factor 20

decreases by factor 10

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Explanation

The velocity of sound v= n $λ$

$\frac{v_{1}}{v_{2}} = \frac{n_{1} λ_{1}}{n_{2} λ_{2}} (but n_{1} = n_{2}) λ_{2} = λ_{1} \frac{v_{2}}{v_{1}} = λ_{1} \times 10 λ_{2} = 10 λ_{1}$

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Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is :

zero

$\frac{2 π}{3}$

$π$

$\frac{π}{6}$

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Explanation

$y_{1} = \frac{A}{2} = A \sin ω t$

$\Rightarrow ω t = 30 °$

$y_{2} = \frac{A}{2} = A \sin (ω t + ϕ)$

$\Rightarrow ω t + ϕ = 150 °$

$∴$ phase difference

$ϕ = 150 ° - 30 °$

= $120 = \frac{2 π}{3} r a d$

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