Tag: speed and acceleration of travelling wave

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Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The  equation of a transverse wave travel on a rope is given   y = 10 sin $\pi$(0.01x - 2.00t) where y and x in cm and t in seconds.The maximum transverse  speed  of a particle in the rope about 

  1. 62.8 cm / s

  2. 75 cm / s

  3. 100 cm / s

  4. 121 cm / s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The wave equation is y = 10 sin(0.01 * pi * x - 2 * pi * t). The transverse velocity is v_y = dy/dt = -10 * 2 * pi * cos(0.01 * pi * x - 2 * pi * t). The maximum transverse speed is 20 * pi = 20 * 3.14 = 62.8 cm/s.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A wave represented by equation $y = 2(mm) \, sin \, [4 \pi (sec^{-1}) t - 2 \pi (m^{-1}) X]$ is superimposed with another wave $y = 2 (mm) sin [4 \pi (sec^{-1}) t + 2 \pi (m^{-1}) x + \pi/3]$ on a tight string.
Phase difference between two particles with are located at $x _1 = 1/7$ and $x _2 = 5/12$ is :

  1. $0$

  2. $\dfrac{5 \pi}{6}$

  3. $\pi$

  4. $\dfrac{5 \pi}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${y _1} = 2\sin \left[ {4\pi t - 2\pi x} \right]$

${y _2} = 2\sin \left[ {4\pi t - 2\pi x} \right]$
$y = {y _1} + {y _2} = 2\left[ {2\sin 4\pi t\,\,\cos 4\pi x} \right]$
$ = 4\sin 4\pi t\cos 4\pi x$
$y = 4\cos 4\pi \sin 4\pi t$
Amp pass$\left| {4\cos 4\pi x} \right| = 4$
$ \Rightarrow \cos 4\pi x =  \pm 1$
$ = 0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},\frac{5}{4},\frac{6}{4}.......$
$\therefore for\,{x _1} = \frac{1}{9}and\,{x _2} = \frac{5}{{12}}$
hence,
phase difference is$\pi$

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A travelling wave on a string is given by $y = A$ $A \sin \left[ \alpha x + \beta t + \cfrac { \pi } { 6 } \right]$ The displacement and velocity of oscillation of a point $\alpha =$ $0.56 / \mathrm { cm } , \beta = 12 / \mathrm { sec }$ $A = 7.5 \mathrm { cm } , x = 1$ $\mathrm { cm }$ and $\mathrm { t } = 1 \mathrm { s }$ is

  1. $4.6 \mathrm { cm } , 46.5 \mathrm { cm } s ^ { - 1 }$

  2. $3.75 \mathrm { cm } , 77.94 \mathrm { cm } \mathrm { s } ^ { - 1 }$

  3. $1.76 \mathrm { cm } , 7.5 \mathrm { cms } ^ { - 1 }$

  4. $7.5 \mathrm { cm } , 75 \mathrm { cm } \mathrm { s } ^ { - 1 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} y=7.5\sin  \left[ { 0.56\alpha +12t+\dfrac { \pi  }{ 6 }  } \right]  \ at\, \, x=1 \ & \, \, t=1 \ y=7.5\sin  \left[ { 0.56+12+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 12.56+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 4\pi +\dfrac { \pi  }{ 6 }  } \right]  \ =3.75\, \, cm \end{array}$

$\therefore$ Option $B$ is correct .

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A sine wave is travelling in a medium. The minimum distance between the two particles. always having same speed is 

  1. $\lambda / 4$

  2. $\lambda / 3$

  3. $\lambda / 2$

  4. $\lambda $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a sine wave particles that are separated by a distance of odd multiple of half

the wave length move with same speed and but in opposite direction. 
The minimum separation is $\frac{\lambda }{2}$
Option C.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A kite flying at a height h meter has r meter of string paid out at a time of t sec . If the kite moves horizontally with constant velocity v meter/sec then the at which the string is paid out is

  1. $\sqrt{\left ( r^2 h^2 \right )v}$ mt/sec

  2. $\dfrac{v\sqrt{\left ( r^2 h^2 \right )}}{r}$ mt/sec

  3. $\dfrac{r\sqrt{\left ( r^2 h^2 \right )}}{v}$ mt/sec

  4. $\dfrac{\sqrt{\left ( r^2 h^2 \right )}}{rv}$ mt/sec
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The vibration of string of length 60 cm fixed at both ends are represented by the equations $ y=4 sin ( \pi x / 15 ) cos ( 96 \pi / t ) $ where x and y are in cm and t in s. the maximum displacement at x=5 cm is 

  1. $ 2 \sqrt 3 cm $

  2. $4 cm$

  3. $zero$

  4. $ 4 \sqrt 2 cm $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given y = 4 sin(πx/15) cos(96π/t) - note this is TYPO: should be cos(96πt). Maximum displacement occurs when cos(96πt) = ±1. At x=5 cm, amplitude = 4 sin(π×5/15) = 4 sin(π/3) = 4 × √3/2 = 2√3 cm.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A uniform string of length 20 m & mass 1 Kg is hung vertically.Find the speed of wave at the mid point of the string :-

  1. 20 m/s

  2. 30 m/s

  3. $ 10 \sqrt {2} m/s $

  4. 10 m/s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Tension at midpoint: The weight of the lower half of the string is (m/2) * g = 0.5 * 10 = 5 N. Linear mass density mu = 1 kg / 20 m = 0.05 kg/m. Wave speed v = sqrt(T / mu) = sqrt(5 / 0.05) = sqrt(100) = 10 m/s.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A standing wave of time period T is set up in string clamped between two rigid supports at t=0 antitode is at its maximum displacement A

  1. The energy of a node is equal to energy of an anitode for the first time at t=T/8

  2. The energy of node and antitode becomes equal after every T/2 second.

  3. the displacement of the particle of antinode at $ t= \frac {T}{8} is \sqrt 2 A $

  4. The displacement of the particle of node is zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A man generates a ssmmetrical pulse in a string by moving his hand up and down. At $t = 0$ the how hand mowes downuard: The pulse travels with speed of 3$\mathrm { m } / \mathrm { s }$ on the string $&$ his hands passe 6 in each secand from the mean position. Then the point on the string at a distance 3$\mathrm { m }$ will reach its topper arreme first time at time t=

  1. 0.25 sec.

  2. 1 sec

  3. $\frac { 13 } { 12 } \mathrm { sec }$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Hand completes 6 oscillations per second, so period T = 1/6 s. At t=0, hand moves downward. Pulse travels at 3 m/s. Point at 3 m reaches upper extreme when pulse arrives and phase is maximum. Time = distance/speed + T/4 = 3/3 + (1/6)/4 = 1 + 1/24 = 25/24 ≈ 0.25 s.