A simple harmonic wavetrain of amplitude $$5\ cm$$ and frequency $$100\ Hz$$ is travelling in the positive $$x-$$ direction with a velocity of $$30\ m/s$$. The displacement velocity and acceleration at $$t=3s$$ of a particle of the medium situated $$100\ cm$$ from the origin are respectively.

$$-4.33\ cm,+1570\ cm/s,171\times 10^{4}\ cm/s^{2}$$

$$-3.44\ cm+1750\ cm/s, 17\times cm/s^{2}$$

$$4.33\ cm+1570\ cm/s,71\times 10^{4}\ cm/s^{2}$$

$$-3.44\ cm,-1750\ cm/s,171\times 10^{4}\ cm/s^{2}$$

Two coherent waves represented by $$y_1 = A sin \left(\dfrac{2 \pi}{\lambda} x_1 - \omega t + \dfrac{\pi}{6} \right)$$ and $$y_2 = A sin \left(\dfrac{2 \pi}{\lambda} x_2 - \omega t + \dfrac{\pi}{6} \right)$$ are superimposed . The two waves will produce

destructive interference at $$(x_1 - x_2) = 11/24 \lambda$$

constructive interference at $$(x_1 - x_2) = 2 \lambda$$

constructive interference at $$(x_1 - x_2) = 23/24 \lambda$$

destructive interference at $$(x_1 - x_2) = 1.5 \lambda$$

Following are equations of four waves: (i) $$y_1 = a sin \omega \left(t - \dfrac{x}{v} \right)$$ (ii) $$y_2 = a cos \omega \left( t = \dfrac{x}{v} \right)$$ (iii) $$z_1 = a sin \omega \left(t - \dfrac{x}{v} \right)$$ (iv) $$z_2 = a cos \omega \left( t = \dfrac{x}{v} \right)$$ Which of the following statements are correct?

On superposition of (iii) and (iv), a transverse stationary wavw will be formed

On superpositin of waves (i) and (iii), a travelling wave having amplitude $$a\sqrt{2}$$ will be formed

Superposition of waves (ii) and (iio) is not possible

On superposition of (i) and (ii), a stationary wave having amplitude $$a\sqrt{2}$$ will be formed

Two separated sources emit sinusoidal travelling waves but have the same wavelength $$ \lambda $$ and are in phase at their respective sources. One travels a distance $$ l_{1} $$ to get to the observation point while the other travels a distance $$ l_{2} $$. The amplitude is minimum at the observation point, if $$ l_{1}-l_{2} $$ is an

odd integral multiple of $$ \lambda / 2 $$

odd integral multiple of $$ \lambda $$

even integral multiple of $$ \lambda $$

odd integral multiple of $$ \lambda / 4 $$

Two waves of nearly same amplitude, same frequency travelling with same velocity are superimposing to give phenomenon of interference, If $$a_1$$ and $$a_2$$ be their respectively amplitudes, $$\omega$$ be the frequency for both, v be the velocity for both and $$\Delta \phi$$ is the phase difference between the two waves then,

both the waves must have been travelling in the same direction and must be coherent.

the resultant intensity varies periodically with time and distance.

the resulting intensity with $$\dfrac{I_min}{I_max} = \left (\dfrac{a_1 - a_2} {a_1 + a_2} \right) ^{2}$$ is obtained

$$I_B = I_1 + I_2 + 2 \sqrt{I_1 I_2} cos (\Delta \phi)$$, where constructive interference is obtained for path difference that are even multiple of $$1/2 \lambda$$.

The amplitude of a wave represented by displacement equation $$y = \dfrac{1}{\sqrt{a}} sin\omega t \pm \dfrac{1}{\sqrt{b}} cos \omega t$$will be

$$\dfrac{\sqrt{a}\pm\sqrt{b}}{ab}$$

A man $$x$$ can hear only upto $$10 kHz$$ and another man y upto $$20 Hz$$. A note of frequency $$500 Hz$$ is produced before them from a stretched string. Then

Both will hear sounds of same pitch and same quality

Both will hear sounds of same pitch but different quality

Both will hear sounds of different pitch but same quality

Both will hear sounds of different pitch and different quality

MCQ Questions for Class 11 Engineering Physics Waves Quiz 13

Two simple harmonic motions are represented by the equations

y_{1} = 10 sin (3 π t + \frac{π}{4})

$y_1=10\sin \left(3\pi t+\dfrac{\pi}{4}\right)$
and

y_{2} = 5 (3 sin 3 π t + \sqrt{3} cos 3 π t)

$y_2=5(3\sin 3\pi t+\sqrt 3 \cos 3\pi t)$ Their amplitudes are in the ratio of :

0%
$\sqrt 3$
0%
$1/\sqrt 3$
0%
$2$
0%
$1/6$

A simple harmonic wavetrain of amplitude

5 c m

$5\ cm$ and frequency

100 H z

$100\ Hz$ is travelling in the positive

x -

$x-$ direction with a velocity of

30 m / s

$30\ m/s$ . The displacement velocity and acceleration at

t = 3 s

$t=3s$ of a particle of the medium situated

100 c m

$100\ cm$ from the origin are respectively.

0%
$-3.44\ cm+1750\ cm/s, 17\times cm/s^{2}$
0%
$4.33\ cm+1570\ cm/s,71\times 10^{4}\ cm/s^{2}$
0%
$-4.33\ cm,+1570\ cm/s,171\times 10^{4}\ cm/s^{2}$
0%
$-3.44\ cm,-1750\ cm/s,171\times 10^{4}\ cm/s^{2}$

Explanation

y = A sin (\frac{2 π μ}{v} (v t - x))

$y=A\sin (\dfrac{2\pi \mu}{v}(vt-x))$

y = 5 sin (\frac{2 π 100}{3000} (3000 \times 3 - 100)) = - 4.45 c m

$y=5\sin (\dfrac{2\pi 100}{3000}(3000\times 3-100))=-4.45cm$

velocity=

\frac{d y}{d x} = A \times 2 π μ cos (\frac{2 π μ}{v} (v t - x)) = 5 \times 2 π \times 100 cos (\frac{2 π \times 100}{3000} (3000 \times 3 - 180)) = 1571 c m / s

$\dfrac{dy}{dx}=A\times 2\pi\mu\cos(\dfrac{2\pi\mu}{v}(vt-x))=5\times 2\pi\times 100\cos(\dfrac{2\pi\times 100}{3000}(3000\times 3-180))=1571cm/s$

accleration=

\frac{d^{2} y}{d x^{2}} = - A \times 4 π^{2} μ^{2} s i n (\frac{2 π μ}{v} (v t - x)) = - 5 \times 4 π^{2} \times 10000 sin (\frac{2 π \times 100}{3000} (3000 \times 3 - 180)) = 171 \times 10^{4} c m / s^{2}

$\dfrac{d^2y}{dx^2}=-A\times 4\pi^2\mu^2sin(\dfrac{2\pi\mu}{v}(vt-x))=-5\times 4\pi^2\times 10000\sin(\dfrac{2\pi\times 100}{3000}(3000\times 3-180))=171\times 10^{4}cm/s^2$

Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is

120^{\circ}

$120^{\circ}$ . The resultant amplitude will be

0%
$A$
0%
$2A$
0%
$4A$
0%
$\sqrt{2 A}$

A wave motion has the function

Y = a_{0} sin (ω t - k x)

$Y=a_{0}\sin (\omega t-kx)$ . The graph in the figure shows how the displacement

y

$y$ at a fixed point varies with time

t

$t$ . Which one of the labeled points shows a displacement equal to that at the position

x = π / 2 k

$x=\pi/2k$ at time

t = 0

$t=0$

0%
$P$
0%
$Q$
0%
$R$
0%
$S$

Explanation

Given ,

Y = a_{0} sin (ω t - k x)

$Y=a_{0}\sin (\omega t-kx)$ .

Now, at

x = \frac{π}{2 k} a n d t = 0

$x=\dfrac{\pi}{2k} \ and\ t=0$

we get,

Y = a_{0} sin (ω (0) - k (\frac{π}{2 k})

$Y=a_{0}\sin (\omega (0)-k(\dfrac{\pi }{2k})$ .

Y = - a_{0}

$Y=-a_0$

from the figure point (Q) shows a displacement equal to (-a_0).

hence, option B is correct

A simple harmonic plane wave propagates along x-axis in a medium. The displacement of the particles as a function of time is shown in figure, for

x = 0

$x = 0$ (curve 1) and

x = 7

$x = 7$ (curve 2).
The two particles are within a span of one wavelength.
The speed of the wave is

0%
$12 \,m/s$
0%
$24 \,m/s$
0%
$8 \,m/s$
0%
$16 \,m/s$

Explanation

Let general wave equation is

y = A s i n (ω t - k x + ϕ)

$y = A \,sin (\omega t - kx + \phi)$

v = \frac{d y}{d t} = A ω c o s (ω t - k x + ϕ)

$v = \dfrac {dy}{dt} = A \omega cos (\omega t - kx + \phi)$
for curve

(1), x = 0

$(1 ) , x = 0$
at

t = 0, x = 0,

$t = 0, x = 0 ,$ we have

y = 0

$y = 0$

\Rightarrow 0 = A s i n [ϕ] \Rightarrow s i n ϕ = 0

$\Rightarrow 0 = A \,sin[\phi] \Rightarrow sin \phi = 0$

\Rightarrow ϕ = 0 o r π

$\Rightarrow \phi = 0 \,or \,\pi$
here

ϕ = π

$\phi = \pi$ (because velocity is negative)
for curve

(2), x = 7 c m

$(2), x = 7 \,cm$
at

t = 0, x = 7 c m, y = - 1

$t = 0, x = 7 \,cm, y = -1$

- 1 = s i n (- k \times 7 + π)

$-1 = sin(-k \times 7 + \pi)$

\Rightarrow s i n (- 7 k + π) = - 1 / 2

$\Rightarrow sin(-7k + \pi) = -1/2$

\Rightarrow - 7 k + π = 2 n π + \frac{7 π}{6}

$\Rightarrow -7k + \pi = 2n\pi + \dfrac {7\pi}{6}$ or

2 n π + \frac{11 π}{6}

$2n\pi + \dfrac {11\pi}{6}$
here

\Rightarrow - 7 k + π = 2 n π + \frac{11 π}{6}

$\Rightarrow -7k + \pi = 2n\pi + \dfrac {11\pi}{6}$
(because at t = 0, velocity is positive)

\Rightarrow - 7 (\frac{2 π}{λ}) = 2 n π + \frac{5 π}{6}

$\Rightarrow -7 \left ( \dfrac {2\pi}{\lambda} \right ) = 2n\pi + \dfrac {5\pi}{6}$

\Rightarrow λ = \frac{- 14 π}{5 π / 6 + 2 n π}

$\Rightarrow \lambda = \dfrac {-14\pi}{5\pi/6 + 2n\pi}$

\Rightarrow λ = \frac{- 84}{12 n + 5}

$\Rightarrow \lambda = \dfrac {-84}{12n + 5}$
for

n = - 1, λ = 12 c m

$n = -1, \lambda = 12 \,cm$
for

n = - 2, λ = \frac{84}{19} c m

$n = -2, \lambda = \dfrac {84}{19}cm$ (not possible)
because

λ > 7 c m

$\lambda > 7 \,cm$

v = f λ = 100 \times \frac{12}{100} = 12 m / s

$v = f\lambda = 100 \times \dfrac {12}{100} = 12 \,m/s$

Two coherent waves represented by

y_{1} = A s i n (\frac{2 π}{λ} x_{1} - ω t + \frac{π}{6})

$y_1 = A sin \left(\dfrac{2 \pi}{\lambda} x_1 - \omega t + \dfrac{\pi}{6} \right)$ and

y_{2} = A s i n (\frac{2 π}{λ} x_{2} - ω t + \frac{π}{6})

$y_2 = A sin \left(\dfrac{2 \pi}{\lambda} x_2 - \omega t + \dfrac{\pi}{6} \right)$ are superimposed . The two waves will produce

0%
constructive interference at $(x_1 - x_2) = 2 \lambda$
0%
constructive interference at $(x_1 - x_2) = 23/24 \lambda$
0%
destructive interference at $(x_1 - x_2) = 1.5 \lambda$
0%
destructive interference at $(x_1 - x_2) = 11/24 \lambda$

Explanation

Path difference can be given as

Δ ϕ = \frac{2 π x_{1}}{λ} - ω t + \frac{π}{4} - 2 π \frac{x_{2}}{λ} + ω t - \frac{π}{6}

$\Delta \phi = \dfrac{2 \pi x_1}{\lambda} - \omega t + \dfrac{\pi}{4} - 2 \pi \dfrac{x_2}{\lambda} + \omega t - \dfrac{\pi}{6}$

= \frac{2 π}{λ} (x_{1} - x_{2}) + \frac{π}{12}

$= \dfrac{2 \pi}{\lambda} (x_1 - x_2) + \dfrac{\pi}{12}$

for constructive interference

\frac{2 π}{λ} (x_{1} - x_{2}) + \frac{π}{12} = 2 n π

$\dfrac{2 \pi}{\lambda} (x_1 - x_2) + \dfrac{\pi}{12} = 2 n \pi$ n = 0, 1, 2,....

for destructive interference

\frac{2 π}{λ} (x_{1} - x_{2}) + \frac{π}{12} = (2 n - 1) π

$\dfrac{2 \pi}{\lambda} (x_1 - x_2) + \dfrac{\pi}{12} = (2n - 1) \pi$

Two vibration strings of the same material but lengths L and 2L have radii 2r and r, respectively. The are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency

n_{1}

$n_1$ and the other with frequency

n_{2}

$n_2$ . The ratio

n_{1} / n_{2}

$n_1/n_2$ is given by

0%
2
0%
4
0%
8
0%
1

Explanation

n_{1} = \frac{1}{2 l} \sqrt{[\frac{T}{4 π r^{2} ρ}]}

$n_{1}=\frac{1}{2l}\sqrt{\left [ \frac{T}{4\pi r^{2}\rho } \right ]}$

and

n_{2} = \frac{1}{4 l} \sqrt{[\frac{T}{π r^{2} ρ}]}

$n_{2}=\frac{1}{4l}\sqrt{\left [ \frac{T}{\pi r^{2}\rho } \right ]}$

∴ \frac{n_{1}}{n_{2}} = 2 \times \frac{1}{2} = 1

$\therefore \frac{n_{1}}{n_{2}}=2\times \frac{1}{2}=1$

Figure 5.55 shows a student setting up wave on a long stretched string. The student's hand makes one complete up and down movement in

0.4 s

$0.4 \,s$ and in each up and down movement the hand moves by a height of

0.3 m .

$0.3 \,m.$ The wavelength of the waves on the string is

0.8 m .

$0.8 \,m.$
The amplitude of the wave is

0%
$0.15 \,m$
0%
$0.3 \,m$
0%
$0.075 \,m$
0%
cannot be predicted

Explanation

F r e q u e n c y = \frac{1}{T i m e P e r i o d} = \frac{1}{0.4} = 2.5 H z

$Frequency = \dfrac {1}{Time \,Period} = \dfrac {1}{0.4} = 2.5 \,Hz$

A m p l i t u d e = \frac{1}{2} \times 0.3 m = 0.15 m

$Amplitude = \dfrac {1}{2} \times 0.3 \,m = 0.15 \,m$

Four pieces of string each of length L are joined end to end to make a long string of length

4 L .

$4L.$ The linear mass density of the strings are

μ, 4 μ, 9 μ

$\mu, 4\mu, 9\mu$ and

16 μ,

$16\mu,$ respectively. One end of the combined string is tied to a fixed support and a transverse wave has been generated at the other end having frequency

f

$f$ (ignore any reflection and absorptions). String has been stretched under a tension

F .

$F.$
Find the ratio of wavelengths of the waves on four strings, starting from right hand side.

0%
$12 : 6 : 4 : 3$
0%
$4 : 3 : 2 : 1$
0%
$3 : 4 : 6 : 12$
0%
$1 : 2 : 3 : 4$

Explanation

Frequency of wave is same on all the four strings. So,

λ_{1} = \frac{v_{1}}{f},

$\lambda_1 = \dfrac {v_1}{f},$

λ_{2} = \frac{v_{2}}{f},

$\lambda_2 = \dfrac {v_2}{f},$

λ_{3} = \frac{v_{3}}{f},

$\lambda_3 = \dfrac {v_3}{f},$

λ_{4} = \frac{v_{4}}{f},

$\lambda_4 = \dfrac {v_4}{f},$

λ_{4} : λ_{3} : λ_{2} : λ_{1} = \frac{1}{4} : \frac{1}{3} : \frac{1}{2} : 1 = 3 : 4 : 6 : 12

$\lambda_4 : \lambda_3 : \lambda_2 : \lambda_1 = \dfrac {1}{4} : \dfrac {1}{3} : \dfrac {1}{2} : 1 = 3 : 4 : 6 : 12$

Following are equations of four waves:
(i)

y_{1} = a s i n ω (t - \frac{x}{v})

$y_1 = a sin \omega \left(t - \dfrac{x}{v} \right)$
(ii)

y_{2} = a c o s ω (t = \frac{x}{v})

$y_2 = a cos \omega \left( t = \dfrac{x}{v} \right)$
(iii)

z_{1} = a s i n ω (t - \frac{x}{v})

$z_1 = a sin \omega \left(t - \dfrac{x}{v} \right)$
(iv)

z_{2} = a c o s ω (t = \frac{x}{v})

$z_2 = a cos \omega \left( t = \dfrac{x}{v} \right)$
Which of the following statements are correct?

0%
On superpositin of waves (i) and (iii), a travelling wave having amplitude $a\sqrt{2}$ will be formed
0%
Superposition of waves (ii) and (iio) is not possible
0%
On superposition of (i) and (ii), a stationary wave having amplitude $a\sqrt{2}$ will be formed
0%
On superposition of (iii) and (iv), a transverse stationary wavw will be formed

Two separated sources emit sinusoidal travelling waves but have the same wavelength

λ

$\lambda$ and are in phase at their respective sources. One travels a distance

l_{1}

$l_{1}$ to get to the observation point while the other travels a distance

l_{2}

$l_{2}$ . The amplitude is minimum at the observation point, if

l_{1} - l_{2}

$l_{1}-l_{2}$ is an

0%
odd integral multiple of $\lambda$
0%
even integral multiple of $\lambda$
0%
odd integral multiple of $\lambda / 2$
0%
odd integral multiple of $\lambda / 4$

Two waves of nearly same amplitude, same frequency travelling with same velocity are superimposing to give phenomenon of interference, If

a_{1}

$a_1$ and

a_{2}

$a_2$ be their respectively amplitudes,

ω

$\omega$ be the frequency for both, v be the velocity for both and

Δ ϕ

$\Delta \phi$ is the phase difference between the two waves then,

0%
the resultant intensity varies periodically with time and distance.
0%
the resulting intensity with $\dfrac{I_min}{I_max} = \left (\dfrac{a_1 - a_2} {a_1 + a_2} \right) ^{2}$ is obtained
0%
both the waves must have been travelling in the same direction and must be coherent.
0%
$I_B = I_1 + I_2 + 2 \sqrt{I_1 I_2} cos (\Delta \phi)$ , where constructive interference is obtained for path difference that are even multiple of $1/2 \lambda$ .

Explanation

This is the case of sustained interference in which position of maxima and minima remains fixed all over the screen.

\frac{I_{m} i n}{I_{m} a x} = {(\frac{a_{1} - a_{2}}{a_{1} + a_{2}})}^{2}

$\dfrac{I_min}{I_max} = \left (\dfrac{a_1 - a_2} {a_1 + a_2} \right) ^{2}$
And both waves must have been travelling in the same direction with a constant phase difference (condition for coherence)

n waves are produced on a string in 1 s. When the radius of the string is doubled and the tension is maintained the same, the number of waves produced in 1 s for the same harmonic will be

0%
2n
0%
$\frac {n} {3}$
0%
$\frac {n} {2}$
0%
$\frac {n} {\sqrt{2}}$

Explanation

d. Given that the frequency of wave produced in the string is l/n.

∴ \frac{1}{n} = \frac{1}{2 π} \sqrt{\frac{T}{m}}

$\therefore \dfrac{1} {n} = \dfrac{1} {2 \pi} \sqrt {\dfrac{T} {m}}$
Now T = 2T

Therefore, new frequency is

f = \frac{1}{2 π} \sqrt{\frac{2 T}{m}} = \sqrt{2} \times \frac{1}{n}

$f = \dfrac {1} {2\pi} \sqrt {\dfrac {2T} {m}} = \sqrt {2} \times \dfrac{1} {n}$

Therefore, the number of waves produced per second is

\frac{1}{f} = \frac{n}{\sqrt{2}}

$\dfrac {1} {f} = \dfrac {n} {\sqrt{2}}$

One end of a 2.4 m string is held fixed and the other end is attached to a weightless ring that can slide along a frictionless rod as shown in Fig. 7.The three longest possible wavelength for standing waves in this string are respectively

0%
4.8 m, 1.6 m and 0.96 m
0%
9.6 m, 3.2 m and 1.92 m
0%
2.4 m, 0.8 m and 0.48 m
0%
1.2 m, 0.4 m and 0.24 m

Explanation

b. When the end of the string is free to move, the string being attached to weightless ring that can slide freely along the rod, the phase of reflected pulse is unchanged antinode is formed at the ring.

l = \frac{λ}{4} \Rightarrow λ_{1} = 4 l = 9.6 m

$l = \frac{\lambda} {4} \Rightarrow \lambda_1 = 4l = 9.6 \space m$

λ_{2} = \frac{4 l}{3} = 3.2 m

$\lambda_2 = \frac {4l} {3} = 3.2 \space m$

λ_{3} = \frac{4 l}{5} = \frac{9.6}{5} = 1.92 m

$\lambda_3 = \frac{4l} {5} = \frac {9.6} {5} = 1.92 \space m$
1647564_1751613_ans_2b4a23c3ab9d400ba0a9bdfb1701b960.PNG

Which of the following travelling wave will produce standing wave, with nodes at x = 0, when superimposed on

y = A sin (ω t - k x)

$y = A \sin{(\omega t - kx)}$

0%
$A \sin{(\omega t + kx)}$
0%
$A \sin{(\omega t + kx + \pi)}$
0%
$A \cos{(\omega t + kx)}$
0%
$A \cos{(\omega t + kx + \pi)}$

Explanation

b. Substituting x = 0, we have given wave

y = A sin ω t

$y = A \sin{\omega t }$ at x = 0 other should have

y = - A sin ω t = A sin (ω + π)

$y = - A \sin{\omega t }=A\sin(\omega +\pi)$ equation so displacement may be zero at all the time. Hence, option (b) is correct.

Microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflector. Between positions of 14 successive maxima, the detector travels a distance 0.14 m. If the velocity of light is

3 \times 10^{8}

$3 \times 10^{8}$ m/s, find the frequency of the transmitter.

0%
$1.5 \times 10^{10} Hz$
0%
$10^{10} Hz$
0%
$3 \times 10^{10} Hz$
0%
$6 \times 10^{10} Hz$

Explanation

a. If detector moves x distance, distance from direct sound increases by x and distance from reflected sound decreases by x so path difference created = 2x

2 (0.14) = 14 λ = 14 c / f

$2(0.14) = 14 \lambda = 14 \space c/f$

f = \frac{14 \times 3 \times 10^{8}}{0.14 \times 2} = 1.5 \times 10^{10} H z

$f = \frac {14 \times 3 \times 10^{8}} {0.14 \times 2} = 1.5 \times 10^{10}Hz$

Let the two waves

y_{1} = A sin (k x - ω t)

$y_1 = A \sin {(kx - \omega t)}$ and

y_{2} = A sin (k x + ω t)

$y_2 = A \sin {(kx + \omega t)}$ form a standing wave on a string. Now if an additional phase difference of

ϕ

$\phi$ is created between two waves, then

0%
the standing wave will have a different frequency
0%
the standing wave will have a different amplitude for a given point
0%
the spacing between two consecutive nodes will change
0%
none of the above

Explanation

b. Initially the standing wave equation is

y_{1} = 2 A sin k x cos ω t

$y_1 = 2A \sin {kx} \cos {\omega t}$

If phase difference

ϕ

$\phi$ is added to one of waves, then resulting standing wave equation is

y = 2 A sin (k x + \frac{ϕ}{2}) cos (ω t - \frac{ϕ}{2})

$y = 2A \sin {\left (kx + \frac{\phi} {2} \right )} \cos {\left (\omega t - \frac{\phi} {2} \right )}$

Here, frequency does not change and also spacing between two successive nodes does not change as its value for both is

π / k

$\pi/k$ . But for a particle, in standing wave, amplitude changes.

Which of the following are transferred from one place to another place by the waves ?

0%
mass
0%
wavelength
0%
velocity
0%
energy

Two waves are given by

y_{1} = a sin (ω t - k x)

$y_1 = a \sin \left (\omega t - kx \right )$ and

y_{2} = a cos (ω t - k x)

$y_2 = a \cos \left (\omega t - kx \right )$ . The phase difference between the two waves is

0%
$\frac{\pi} {4}$
0%
$\pi$
0%
$\frac{\pi} {8}$
0%
$\frac{\pi} {2}$

Explanation

The two equations of the waves are:

y_{1} = a sin (ω t - k x)

$y_1 = a \sin \left (\omega t - kx \right )$

and

y_{2} = a cos (ω t - k x)

$y_2 = a \cos \left (\omega t - kx \right )$

The second equation can be represented as:

= a sin (ω t - k x + \frac{π}{2})

$= a \sin \left (\omega t - kx + \dfrac{\pi} {2} \right )$

Hence phase difference between these two is

\frac{π}{2}

$\dfrac{\pi} {2}$

If two waves having amplitudes

2 A

$2A$ and

A

$A$ and same frequency andvelocity, propagate in the same direction in the same phase, the resulting amplitude will be

0%
$3A$
0%
$\sqrt{5} A$
0%
$\sqrt{2} A$
0%
$A$

Explanation

In the same phase

ϕ = 0

$\phi = 0$ so resultant amplitude

= a_{1} + a_{2} = 2 A + A = 3 A

$= a_1 + a_2 = 2A + A = 3A$

If amplitude of waves at distance r from a point source is A, the amplitude at a distance 2r will be

0%
$2A$
0%
$A$
0%
$A/2$
0%
$A/4$

Explanation

We know that the relation between the amplitude and the distance is given as:

I \propto a^{2} \propto \frac{1}{d^{2}}

$I \propto a^{2} \propto \dfrac{1} {d^{2}}$

Therefore, the amplitiude at a distance of

2 r

$2r$ will be:

\Rightarrow a \propto \frac{1}{d}

$\Rightarrow a \propto \dfrac{1} {d}$

\frac{A_{1}}{A_{2}} = \frac{r_{2}}{r_{1}}

$\dfrac{A_1}{A_2}=\dfrac{r_2}{r_1}$

A_{2} = \frac{r}{2 r}

$A_2=\dfrac{r}{2r}$

A_{2} = \frac{A}{2}

$A_2=\dfrac{A}{2}$

Two waves of same frequency and intensity superimpose with each other in opposite phases, then after superposition the

0%
Intensity increases by 4 times
0%
Intensity increases by two times
0%
Frequency increases by 4 times
0%
None of these

Two waves are propagating to the point

P

$P$ along a straight line produced by two sources

A

$A$ and

B

$B$ of simple harmonic and of equal frequency. The amplitude of every wave at

P

$P$ is

a

$a$ and the phase of A is ahead by

\frac{p i}{3}

$\dfrac{pi}{3}$ than that of

B

$B$ and the distance

A P

$AP$ is greater than

B P

$BP$ by 50

c m

$cm$ . Then the resultant amplitude at the point

P

$P$ will be, if the wavelength is 1 meter

[BVP 2003]

0%
$2a$
0%
$\sqrt{3}$
0%
$a\sqrt{2}$
0%
$a$

Explanation

Path difference

(△ x) = 50 c m = \frac{1}{2} m

$(\triangle x) = 50 cm = \dfrac{1}{2}m$

∴

$\therefore$ Phase difference

△ ϕ = \frac{2 π}{λ} △ x \Rightarrow ϕ = \frac{2 p i}{1} \frac{1}{2} = π

$\triangle \phi = \dfrac{2 \pi}{\lambda}\triangle x \Rightarrow \phi = \dfrac{2pi}{1} \dfrac{1}{2} = \pi$
Total phase difference

= π - \frac{π}{3} = \frac{2 π}{3}

$= \pi - \dfrac{\pi}{3} = \dfrac{2 \pi}{3}$

\Rightarrow A = \sqrt{a^{2} + a^{2} + 2 a^{2} c o s (2 π / 3)} = a

$\Rightarrow A = \sqrt{a^2 + a^2 + 2a^2 cos(2 \pi / 3)} = a$

In a plane progressive wave given by

y = 25 cos (2 π t - π x)

$y= 25 \cos \left (2\pi t - \pi x \right )$ , the amplitude and frequency are respectively [BCECE 2003]

0%
25, 100
0%
25, 1
0%
25, 2
0%
$50 \pi$ , 2

Explanation

Comparing the given equation with

y = a cos (ω t - k x)

$y = a \cos \left ( \omega t - kx \right )$

a = 25, ω = 2 π t = 2 π \Rightarrow n = 1

$a = 25, \omega = 2\pi t = 2\pi \Rightarrow n = 1$ Hz

A simple harmonic progressive wave is represented by the equation :

y = 8 sin 2 π (0.1 x - 2 t)

$y = 8 \sin 2 \pi \left (0.1x - 2t \right )$ where x and y are in cm and t is in seconds. At any instant the phase difference between two particles separated by 2.0 cm in the x-direction is [MP PMT 2000]

0%
18
0%
36
0%
54
0%
72

Explanation

From the given equation

k = 0.2 π

$k = 0.2 \pi$

\Rightarrow \frac{2 π}{λ} = 0.2 π \Rightarrow λ = 10 c m

$\Rightarrow \frac{2 \pi} {\lambda} = 0.2 \pi \Rightarrow \lambda = 10 \space cm$

Δ ϕ = \frac{2 π}{λ} Δ x = \frac{2 π}{10} \times 2 = \frac{2 π}{5} = 72^{0}

$\Delta \phi = \frac{2 \pi} {\lambda} \Delta x = \frac{2 \pi} {10} \times 2 = \frac{2 \pi} {5} = 72^{0}$

The displacement of the interfering light waves are

y_{1} = 4 s i n ω t

$y_1 = 4 sin\omega t$ and

y_{2} = 3 s i n (ω t + \frac{π}{2})

$y_2=3 sin \left(\omega t + \dfrac{\pi}{2}\right)$ . What is the amplitude of the resultant wave

0%
$5$
0%
$7$
0%
$1$
0%
$0$

A transverse progressive wave on a stretched string has a velocity of

10 m s^{- 1}

$10ms^{-1}$ and a frequency of 100 Hz. The phase difference between two particles of the string which are 23 cm apart will be

0%
$\frac{\pi} {8}$
0%
$\frac{\pi} {4}$
0%
$\frac{3\pi} {8}$
0%
$\frac{\pi} {2}$

Explanation

The phase difference of a wave is related to its path difference by the following relation,

△ ϕ = \frac{2 π}{λ} △ x

$\triangle \phi = \dfrac{2\pi}{\lambda} \triangle x$ .....(i)

where

△ ϕ

$\triangle\phi$ is the phase difference,

△ x

$\triangle x$ is the path difference and

λ

$\lambda$ is the wavelength.

In the question the velocity of the wave is given as

v = 10 m / s

$v=10 m/s$ and the frequency is given as

ν = 100 H z

$\nu =100 Hz$ .

The wavelength is given

v = ν λ

$v=\nu \lambda$

or,

λ = \frac{v}{ν}

$\lambda = \dfrac{v}{\nu}$

or,

λ = 0.1 m

$\lambda = 0.1m$

Using the value of wavelength in equation (i), we get,

△ ϕ = \frac{2 π}{0.2} \times 2.3 \times \times 10^{- 2}

$\triangle \phi=\dfrac{2\pi}{0.2} \times 2.3 \times \times 10^{-2}$

△ ϕ = \frac{π}{2}

$\triangle \phi = \dfrac{\pi}{2}$ which is our required phase difference.

The correct answer is option (D).

Two waves of frequencies

20 H z

$20 Hz$ and

30 H z

$30 Hz$ . Travels out from a common point. The phase difference between them after

0.6

$0.6$ sec is

0%
$12\pi$
0%
$\dfrac{\pi} {2}$
0%
$\pi$
0%
$\dfrac{3\pi} {4}$

Two waves

y_{1} = A_{1} s i n (ω t - β_{1}) y_{2} = A_{2} s i n (ω t - β_{2})

$y_1 = A_1 sin(\omega t -\beta_1) y_2 = A_2 sin(\omega t - \beta_2)$ Superimpose to form a resultant wave whose amplitude is [CPMT 1999]

0%
$\sqrt{A_1^2 + A_2^2 +2A_1A_2cos(\beta_1 - \beta_2) }$
0%
$\sqrt{A_1^2 + A_2^2 +2A_1A_2sin(\beta_1 - \beta_2) }$
0%
$A_1 + A_2$
0%
$| A_1 + A_2 |$

Explanation

Phase difference between the two waves is

ϕ = (ω t - β_{2}) = (ω t - β_{1}) = (β_{1} - β_{2})

$\phi = (\omega t - \beta_2) = (\omega t - \beta_1) = (\beta_1 -\beta_2)$

∴

$\therefore$
Resultant amplitude

A = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} c o s (β_{1} - β_{2})}

$A = \sqrt {A_1^2 +A_2^2 + 2A_1A_2 cos(\beta_1 - \beta_2 )}$

The amplitude of a wave represented by displacement equation

y = \frac{1}{\sqrt{a}} s i n ω t \pm \frac{1}{\sqrt{b}} c o s ω t

$y = \dfrac{1}{\sqrt{a}} sin\omega t \pm \dfrac{1}{\sqrt{b}} cos \omega t$ will be

0%
$\dfrac{a+b}{ab}$
0%
$\dfrac{\sqrt{a}+{b}}{ab}$
0%
$\dfrac{\sqrt{a}\pm\sqrt{b}}{ab}$
0%
$\sqrt{\dfrac{a+b}{ab}}$

Explanation

y = \frac{1}{\sqrt{a}} s i n ω t \pm \frac{1}{\sqrt{b}} s i n (ω t + \frac{π}{2})

$y = \dfrac{1}{\sqrt{a}} sin \omega t \pm \dfrac{1}{\sqrt{b}}sin \left(\omega t + \dfrac{\pi}{2}\right)$
Here phase differnce =

\frac{π}{2} ∴

$\dfrac{\pi}{2}\therefore$ The resultant amplitude

\sqrt{{(\frac{1}{\sqrt{a}})}^{2} + {(\frac{1}{\sqrt{b}})}^{2}} = \sqrt{\frac{1}{a} + \frac{1}{b}} = \sqrt{\frac{a + b}{a b}}

$\sqrt { \left( \dfrac { { 1 } }{ \sqrt { a } } \right)^2 +\left( \dfrac { 1 }{ \sqrt { b } } \right) ^2 } =\sqrt { \dfrac { 1 }{ a } +\dfrac { 1 }{ b } } =\sqrt { \dfrac { a+b }{ ab } }$

The path difference between the two waves

y_{1} = a_{1} sin (ω t - \frac{2 π x}{λ})

$y_1 = a_1 \sin \left (\omega t - \frac{2 \pi x} {\lambda} \right )$ and

y_{2} = a_{2} cos (ω t - \frac{2 π x}{λ} + ϕ)

$y_2 = a_2 \cos \left (\omega t - \frac{2 \pi x} {\lambda} + \phi \right )$ [MP PMT 1994]

0%
$\frac{\lambda} {2 \pi} \phi$
0%
$\frac{\lambda} {2 \pi} \left (\phi + \frac{\pi} {2} \right )$
0%
$\frac {2 \pi} {\lambda} \left (\phi - \frac{\pi} {2} \right )$
0%
$\frac {2 \pi} {\lambda}\phi$

Explanation

y_{1} = a_{1} sin (ω t - \frac{2 π x}{λ})

$y_1 = a_1 \sin \left (\omega t - \frac{2 \pi x} {\lambda} \right )$

y_{2} = a_{2} cos (ω t - \frac{2 π x}{λ} + ϕ) = a_{2} sin (ω t - \frac{2 π x}{λ} + ϕ + \frac{π}{2})

$y_2 = a_2 \cos \left (\omega t - \frac{2 \pi x} {\lambda} + \phi \right ) = a_2 \sin \left (\omega t - \frac{2 \pi x} {\lambda} + \phi + \frac{\pi} {2}\right )$
So phase difference =

ϕ + \frac{π}{2}

$\phi + \frac{\pi} {2}$ and

Δ = \frac{λ}{2 π} (ϕ + \frac{π}{2})

$\Delta = \frac{\lambda} {2 \pi} \left (\phi + \frac{\pi} {2} \right )$

Two waves are represented by

y_{1} = a s i n (ω t + \frac{π}{6})

$y_1 = a sin \left(\omega t +\dfrac{\pi}{6}\right)$ and

y_{2} = a c o s ω t

$y_2 = a cos \omega t$ What will be their resultant amplitude

0%
$a$
0%
$\sqrt{2}a$
0%
$\sqrt{3} a$
0%
$2 a$

Explanation

A = \sqrt{(a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} c o s ϕ)}

$A = \sqrt{(a_1^2 + a_2^2 + 2a_1a_2 cos\phi)}$
Putting

a_{1} = a_{2} = a

$a_1 = a_2 = a$ and

ϕ = \frac{π}{3}

$\phi = \dfrac{\pi}{3}$ we get,

A = \sqrt{3} a

$A = \sqrt {3}a$

The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is

90^{0}

$90^{0}$ . Then the velocity of wave will be

0%
192 m/s
0%
360 m/s
0%
710 m/s
0%
384 m/s

Explanation

Path difference

Δ = \frac{λ}{2 π} \times ϕ = \frac{λ}{2 π} \times \frac{π}{2} = \frac{λ}{4}

$\Delta = \dfrac{\lambda} {2 \pi} \times \phi = \frac{\lambda} {2 \pi} \times \dfrac{\pi} {2} = \dfrac{\lambda} {4}$

∴ Δ = 0.8 m \Rightarrow \frac{λ}{4} \Rightarrow λ = 3.2 m

$\therefore \Delta = 0.8 \space m \Rightarrow \dfrac{\lambda} {4 } \Rightarrow \lambda = 3.2 \space m$

∴ n λ = 120 \times 3.2 = 384 m / s

$\therefore n\lambda = 120 \times 3.2 = 384 \space m/s$

The phase difference between two waves represented by

y_{1} = 10^{- 6} sin [100 t + (x / 50) + 0.5] m

$y_1 = 10^{-6} \sin \left [100 t + (x/50) + 0.5 \right ] m$ ,

y_{2} = 10^{- 6} cos [100 t + (x / 50)] m

$y_2 = 10^{-6} \cos \left [100 t + (x/50) \right ] m$ where x is expressed in meters and t is expressed in seconds, is approximately [CBSE PMT 2004]

0%
1.5 rad
0%
1.07 rad
0%
2.07 rad
0%
0.6 rad

Explanation

y_{1} = 10^{- 6} sin [100 t + (x / 50) + 0.5] m

$y_1 = 10^{-6} \sin \left [100 t + (x/50) + 0.5 \right ] m$

y_{2} = 10^{- 6} sin [100 t + (\frac{x}{50}) + (\frac{π}{2})] m

$y_2 = 10^{-6} \sin \left [100 t + (\frac{x} {50})+ (\frac{\pi} {2}) \right ] m$
Phase difference

ϕ

$\phi$

= [100 t + (x / 50) + 1.57] - [100 t + (x / 50) + 0.5]

$= \left [100 t + (x/50) + 1.57 \right ] - \left [100 t + (x/50) + 0.5 \right ]$

= 1.07

$= 1.07$ radians

A man

x

$x$ can hear only upto

10 k H z

$10 kHz$ and another man y upto

20 H z

$20 Hz$ . A note of frequency

500 H z

$500 Hz$ is produced before them from a stretched string. Then

0%
Both will hear sounds of same pitch but different quality
0%
Both will hear sounds of different pitch but same quality
0%
Both will hear sounds of different pitch and different quality
0%
Both will hear sounds of same pitch and same quality

The phase difference between the two particles situated on both the side of a node is

0%
$0^\cdot$
0%
$90^\cdot$
0%
$180^\cdot$
0%
$360^\cdot$

Explanation

We know that a node is formed when there is a stationary wave. A node is a point on the wave where the amplitude is always zero, which means that a nodal point never vibrates.

When the two particles are in the two sides of a node their phase directions are completely opposite to each other. This means that the phase difference between them should be

π

$\pi$ or

180^{\circ}

$180^{\circ}$ .

Hence the correct answer is option (C).

The equation

y = A c o s^{2} (2 π n t - 2 π \frac{x}{λ})

$y = A cos^2 \left(2\pi n t - 2\pi \dfrac{x}{\lambda}\right)$ represents a wave with

0%
Amplitude $A/2$ , frequency $2n$ and wavelength $\lambda / 2$
0%
Amplitude $A/2$ , frequency $2n$ and wavelength $\lambda$
0%
Amplitude $A,$ frequency $2n$ and wavelength $2\lambda$
0%
Amplitude $A$ , frequency $n$ and wavelength $\lambda$

Explanation

The given equation can be x written as

y = \frac{A}{2} c o s (4 π n t \frac{4 π x}{λ}) + \frac{A}{2}

$y = \dfrac{A}{2} cos \left( 4\pi n t \dfrac{4\pi x}{\lambda}\right) + \dfrac{A}{2}$

(∴ c o s^{2} θ = \frac{1 + c o s 2 θ}{2})

$\left(\therefore cos^2 \theta = \dfrac{1 + cos 2\theta}{2}\right)$
Hence amplitude =

\frac{A}{2}

$\dfrac{A}{2}$ and frequency =

\frac{ω}{2 π} = \frac{4 π n}{2 π} = 2 n

$\dfrac{\omega}{2\pi} = \dfrac{4\pi n}{2\pi} = 2n$
and wave lenght =

\frac{π}{K} = \frac{2 π}{4 π / λ} = \frac{λ}{2}

$\dfrac{\pi}{K} = \dfrac{2\pi}{4\pi / \lambda} = \dfrac{\lambda}{2}$

Three waves of equal frequency having amplitudes

10 μ m, 4 μ m

$10 \mu m, 4 \mu m$ and

7 μ m

$7 \mu m$ arrive at a given point with successive phase difference of

\frac{π}{2}

$\dfrac{\pi}{2}$ The amplitude of the resulting wave in

μ m

$\mu m$ is given by

0%
7
0%
6
0%
5
0%
4

Explanation

The wave 1 and 3 reach out of phase. Hence resultant phase difference between them is

π

$\pi$

∴

$\therefore$

Resultant amplitude of 1 and 3 = 10 - 7 = 3

μ m

$\mu m$

This wave has phase difference of

\frac{π}{2}

$\dfrac{\pi}{2}$ with

4 μ m

$4 \mu m$

Two waves having sinusoidal waveforms have different wavelengths and different amplitude. They will be having

0%
Same pitch and different intensity
0%
Same quality and different intensity
0%
Different quality and different intensity
0%
Same quality and different pitch

In a wave, the path difference corresponding to a phase difference of

ϕ

$\phi$ is

0%
$\dfrac{\pi}{2 \lambda} \phi$
0%
$\dfrac{\pi}{ \lambda} \phi$
0%
$\dfrac{\lambda}{2\pi} \phi$
0%
$\dfrac{\lambda}{\pi} \phi$

Explanation

For

2 π

$2 \pi$ phase difference

\to

$\rightarrow$ Path difference is

λ

$\lambda$

∴

$\therefore$ For

ϕ

$\phi$ phase difference

\to

$\rightarrow$ Path difference is

\frac{λ}{2 π} ϕ

$\dfrac{\lambda}{2 \pi} \phi$

Equation of motion in the same direction are given by

y_{1} = 2 a sin (ω t - k x)

$y_1 = 2a \sin \left (\omega t - kx \right )$ and

y_{1} = 2 a sin (ω t - k x - θ)

$y_1 = 2a \sin \left (\omega t - kx - \theta \right )$
The amplitude of the medium particle will be [CPMT 2004]

0%
$2a \cos \theta$
0%
$\sqrt{2}a \cos \theta$
0%
$2a \cos \theta/2$
0%
$\sqrt{2}a \cos \theta/2$

Explanation

When two waves are superpositioned, their resultant amplitude is given by,

A_{n e w} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} c o s ϕ}

$A_{new}=\sqrt{A_1^{2}+A_2^2+2A_1 A_2 cos \phi}$ , where

ϕ

$\phi$ is the phase difference.

For these two waves we have

A_{1} = A_{2} = a

$A_1=A_2=a$ , and the phase difference is

θ

$\theta$

∴ a_{n e w} = \sqrt{a^{2} + a^{2} + 2 a c o s θ}

$\therefore a_{new}=\sqrt{a^2+a^2+2acos\theta}$

or,

a_{n e w} = \sqrt{2 a^{2} (1 + c o s θ)}

$a_{new}=\sqrt{2a^2(1+cos\theta)}$

or,

a_{n e w} = \sqrt{2} a \sqrt{1 + c o s θ}

$a_{new}=\sqrt{2} a\sqrt{1+cos\theta}$

We know that

1 + c o s θ = 2 c o s^{2} \frac{θ}{2}

$1+cos\theta = 2cos^2 \dfrac{\theta}{2}$ . So, using this,

a_{n e w} = \sqrt{2} a \sqrt{(2 c o s^{2} \frac{θ}{2})}

$a_{new}=\sqrt{2} a \sqrt{(2cos^2\dfrac{\theta}{2})}$

or,

A_{n e w} = 2 a c o s \frac{θ}{2}

$A_{new}=2acos\dfrac{\theta}{2}$ , which is our required answer.

Given in the graph above, the points

A, B, C, D

$A, B, C, D$ represents state of vibration of a sound wave. From the below-mentioned options which represent the wavelength.

0%
Distance between $A$ and $C$ .
0%
Distance between $A$ and $D$ .
0%
Distance between $A$ and $B$ .
0%
Distance between $B$ and $C$ .

Explanation

A) Distance between

A

$A$ and

C

$C$ .

Light travels in the form of

0%
Waves
0%
Packets
0%
Straight Lines
0%
None of these

A certain transverse sinusoidal wave of wavelength

20 c m

$20 cm$ is moving in the positive

x

$x$ direction. The transverse velocity of the particle at

x = 0

$x = 0$ as a function of time is shown. The amplitude of the motion is :

0%
$\dfrac{5}{\pi} cm$
0%
$\dfrac{\pi}{2}cm$
0%
$\dfrac{10}{\pi}cm$
0%
$2\pi cm$

Explanation

V_{m a x} = A ω = 5 c m / s;

$V_{max}=A\omega =5cm/s;$

T = 4 s e c \Rightarrow ω = \frac{2 π}{4} = \frac{π}{2}

$T=4 sec \Rightarrow \omega = \dfrac{2\pi}{4}=\dfrac{\pi}{2}$

\Rightarrow A = \frac{5}{π / 2} = \frac{10}{π}

$\Rightarrow A =\dfrac{5}{\pi /2} = \dfrac{10}{\pi }$ cm

A wave of frequency 500 Hz has a phase velocity of 360 m/s. The phase difference between the two displacements at a certain point in a time interval of 10

^{- 3}

$^{-3}$ seconds will be how much?

0%
$\displaystyle \frac{\pi}{2}$ radian
0%
$\pi$ radian
0%
$\displaystyle \frac{\pi}{4}$ radian
0%
$\displaystyle \frac{\pi}{8}$ radian

Explanation

According to the concept of simple harmonic wave,

y = a s i n \frac{2 π}{λ} (v t - ϕ)

$y = a sin \dfrac{2 \pi }{ \lambda } (vt - \phi )$

Now, phase angle at point "

x

$x$ " for time"

t_{1}

$t_1$ "

t_{1} = \frac{2 π}{λ} (v t - x)

$t_{1} =\dfrac{2 \pi }{ \lambda } (vt -x)$

now for time "

t_{2}

$t_2$ "

t_{2} = \frac{2 π}{λ} (v t - x)

$t_{2} =\dfrac{2 \pi }{ \lambda } (vt -x)$

phase difference ,

(ϕ_{2} - ϕ_{1}) = \frac{2 π}{λ} (v t_{2} - x) - \frac{2 π}{λ} (v t_{1} - x)

$( \phi _{2} - \phi _{1} )=\dfrac{2 \pi }{ \lambda } (v t_{2} -x) -\dfrac{2 \pi }{ \lambda } (v t_{1} -x)$

= \frac{2 π}{λ} v (t_{2} - t_{1})

$=\dfrac{2 \pi }{ \lambda }v (t_{2} - t_{1})$

= \frac{2 π n λ}{λ} (t_{2} - t_{1})

$= \dfrac{2 \pi n \lambda}{ \lambda } (t_{2}-t_{1})$

= 2 π n (t_{2} - t_{1})

$=2 \pi n(t_{2}-t_{1})$

= 2 π \times 500 \times 10^{- 3}

$=2 \pi \times 500 \times 10^{-3}$

= π r a d i a n

$= \pi radian$

Find the size of object which can be featured with

5 M H z

$5\space MHz$ in water.

0%
0.148 mm
0%
0.3 mm
0%
0.5 mm
0%
0.1 mm

Explanation

v = 1.48 \times 10^{5} \times 10^{- 2} = 1480 m s^{- 1}

$\quad v = 1.48\times10^5\times10^{-2} = 1480\space ms^{-1}$

λ = \frac{v}{f} = \frac{1480}{5 \times 10^{6}} = 0.3 m m

$\quad \lambda = \displaystyle\frac{v}{f} = \displaystyle\frac{1480}{5\times10^6} = 0.3\space mm$

The frequency of fork is 512 Hz and the sound produced by it travels 42 metres as the tuning fork completes 64 vibrations. Find the velocity of sound :

0%
336 m/sec
0%
320 m/sec
0%
340 m/sec
0%
350 m/sec

Explanation

The distance traveled for one vibration is equal to the wavelength. So

64 λ = 42

$64\lambda= 42$

∴ λ = \frac{42}{64} = 0.656

$\therefore \lambda = \dfrac{42}{64} = 0.656$

Velocity of sound

v = λ ν = 0.656 \times 512 = 336 m / s

$v=\lambda \nu=0.656 \times 512=336\: m/s$

A particle is executing SHM of amplitude

A,

$A,$ about the mean position

x = 0.

$x=0.$ Which of the following is a possible phase difference between the positions of the particle at

x = + \frac{A}{2}

$x=+\dfrac{A}{2}$ and

x = - \frac{A}{\sqrt{2}}

$x=-\dfrac{A}{\sqrt{2}}$ .

0%
${75^{\circ}}$
0%
${165^{\circ}}$
0%
${135^{\circ}}$
0%
${195^{\circ}}$

Explanation

The equation for SHM $x=A\cos(\omega t+\phi)$
When $x=\dfrac{A}{2}, \Rightarrow \dfrac{A}{2}=A\cos(\omega t+\phi_1)$
or $60^o=\omega t+\phi_1 ....(1)$
When $x=-\dfrac{A}{\sqrt 2}, \Rightarrow -\dfrac{A}{\sqrt 2}=A\cos(\omega t+\phi_1)$
or $225^o=\omega t+\phi_2 ....(2)$
$(2)-(1) \Rightarrow \phi_2-\phi_1=225-60=165^o$
also $x=A\sin(\omega t+\phi)$
When $x=\dfrac{A}{2}, \Rightarrow \dfrac{A}{2}=A\sin(\omega t+\phi_1)$
or $30^o=\omega t+\phi_1 ....(1)$
When $x=-\dfrac{A}{\sqrt 2}, \Rightarrow -\dfrac{A}{\sqrt 2}=A\sin(\omega t+\phi_1)$
or $225^o=\omega t+\phi_2 ....(2)$
$(2)-(1) \Rightarrow \phi_2-\phi_1=225-30=195^o$

The theory that can explain the phenomenon of interference, diffraction and polarisation is

0%
Wave Theory
0%
Plank's Theory
0%
Wave theory of Light
0%
None of these

Travelling wave travels in medium '1' and enters into another medium '2' in which it's speed gets decreased to

25 %

$25\%$ . Then magnitude of ratio of amplitude of transmitted to reflected wave is

0%
$\dfrac{6}{5}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{1}{7}$
0%
$\dfrac{5}{9}$

Explanation

${V}_{2}=\frac { { V }_{ 1 } }{ 4 } \\ \dfrac { { A }_{ t } }{ { A }_{ r } } =\dfrac { (\dfrac { 2{ V }_{ 2 } }{ { V }_{ 2 }+{ V }_{ 1 } } ) }{ (\dfrac { { V }_{ 2 }-{ V }_{ 1 } }{ { V }_{ 2 }+{ V }_{ 1 } } ) } =(\dfrac { 2{ V }_{ 2 } }{ { V }_{ 2 }-{ V }_{ 1 } } )\\ (\dfrac { 2\dfrac { { V }_{ 1 } }{ 4 } }{ \dfrac { { V }_{ 1 } }{ 4 } -{ V }_{ 1 } } )=\dfrac { 2\dfrac { { V }_{ 1 } }{ 4 } }{ 3\dfrac { { V }_{ 1 } }{ 4 } } =\dfrac { 2 }{ 3 }$

CBSE Questions for Class 11 Engineering Physics Waves Quiz 13 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Waves Quiz 13 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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