Tag: wave motion

Questions Related to wave motion

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45 Hz$. The mass of the wire is $3.5 \times 10^{-2}kg$ and its linear mass density is $4.0 \times 10^{-2} kgm^{-1}$. What is the speed of a transverse wave on the wire?

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

  1. $69 \ ms^{-1}$

  2. $79 \ ms^{-1}$

  3. $89 \ ms^{-1}$

  4. $99 \ ms^{-1}$

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Correct Option: B

A person observe two points on a string as a travelling wave passes them. The points are at $x _ { 1 } = 0$ and $x _2 = 1m$. The transverse motions of the two points are found to be as follows: $y _ { 1 } = 0.2 \sin 3 \pi t$
$y _ { 2 } = 0.2 \sin ( 3 \pi t + \pi/8 )$ What is the frequency in Hertz?

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

  1. $1.5 Hz$

  2. $3 Hz$

  3. $4.5 Hz$

  4. $1 Hz$

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Correct Option: A

A stretched string resonates with tuning fork frequency $512\ Hz$ When of the string is $0.5\ $. The length of the string required to vibrate resonantly with a tuning fork of frequency $256\ Hz$ would  be

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

  1. $0.25\ m$

  2. $0.5\ m$

  3. $1\ m$

  4. $2\ m$

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Correct Option: A

If $n,2n,3n$ are the fundamental frequencies of the three segments into which a string is divided by placing required number of bridges below it. If $n _0$ is the fundamental frequency of the string, then 

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

  1. $n _0=3n$

  2. $n _0=6n$

  3. $n _0=\dfrac{3n}{5}$

  4. $n _0=\dfrac{6n}{11}$

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Correct Option: D

A spring of force constant K is first stretched by distance a from its natural length and then future by distance b. The work done in stretching the part b is

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

  1. $\dfrac{1}{2}$Ka(a-b)

  2. $\dfrac{1}{2}$Ka(a+b)

  3. $\dfrac{1}{2}$Kb(a-b)

  4. $\dfrac{1}{2}$Kb(2a+b)

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Correct Option: D
Explanation:

Work done by spring in its natural length$=\cfrac{1}{2} \times k \times x^{2}= \cfrac{1}{2} \times k \times a^{2}$

So, total work$=\cfrac{1}{2}k(a+b)^{2}$
for work done for stretching 'b'
$\cfrac { 1 }{ 2 } \times k\times (a+b)^{ 2 }-\cfrac { 1 }{ 2 } \times k\times a^{ 2 }=\cfrac { 1 }{ 2 } \times k\times b\times (2a+b)$

Two tuning forks when sounded together produce 5 beat per second. The first tuning fork is in resonance with 16.0 cm wire of a sonometer and the second is in resonace with 16.2 cm wire of the same sonometer. The frequencies of the tuning forks are

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

  1. 100 Hz,105 Hz

  2. 20 Hz,205 Hz

  3. 300 Hz,305 Hz

  4. 400 Hz,405 Hz

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Correct Option: C

The equation of wave in string is $\displaystyle y = 20\sin \frac{\pi x}{2} \cos 40\pi t$ in metre. The speed of the wave is 

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

  1. $Zero$

  2. $80\, m/s$

  3. $320\, m/s$

  4. $160\, m/s$

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Correct Option: B
Explanation:

$\large \begin{array}{l} Here, \ y=20\sin  \frac { { \pi x } }{ 2 } \cos  40\pi t-----(i) \ compare\, with\, eqution,\, (i)\,  \ \Rightarrow y=2r\, \, \sin  \frac { { 2\pi  } }{ \lambda  } \, \times \, \, \cos  \frac { { 2\pi  } }{ \lambda  } Vt \ Now, \ \Rightarrow \frac { { 2\pi  } }{ \lambda  } =\frac { \pi  }{ 2 } \, \, and\, \, \frac { { 2\pi  } }{ \lambda  } V=40\pi  \ so, \ \Rightarrow \frac { \pi  }{ 2 } \, \times V=40\pi  \ \therefore \, \, V=80\, m/s \end{array}$

A body of mass m is tied to one of a spring and whirled round in a horizontal plane with a constant angular velocity and elongation in the spring is 1 cm. If the angular velocity is doubled, the elongation in the spring becomes 5 cm. The original length of spring is 

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

  1. 20 cm

  2. 25 cm

  3. 10 cm

  4. 15 cm

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Correct Option: D

A 100 Hz sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of $3.5\, \times\, 10^{-3}\, kg/m$ and a tension of 35 N. At time t = 0, the point x = 0, has maximum displacement in the positive y direction. Next when this point has zero displacement the slope of the string is $\pi /20$. which of the following expression represent (s) the displacement of string as a function of x (in metre) and t (in second).

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

  1. $y\, =\, 0.025\, cos\, (200 \pi t\, -\, 2 \pi x)$

  2. $y\, =\, 0.5\, cos\, (200 \pi t\, -\, 2 \pi x)$

  3. $y\, =\, 0.025\, cos\, (100 \pi t\, -\, 10 \pi x)$

  4. $y\, =\, 0.5\, cos\, (100 \pi t\, -\, 10 \pi x)$

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Correct Option: A
Explanation:
Let the wave have the form $y=Asin(\omega t-kx+\phi)$
Since the frequency is $100Hz$, $\omega=2\pi\nu=200\pi$
Speed of the wave=$\sqrt{\dfrac{T}{\mu}}=\dfrac{\omega}{k}$
$\implies k=2\pi$
Since displacement is maximum at (x,t)=(0,0), $sin(0+0+\phi)=1$
$\implies \phi=\dfrac{\pi}{2}$
Thus the wave is $y=Acos(\omega t-kx)$
$Slope=\left|\dfrac{dy}{dx}\right|=Aksin(\omega t-kx)=\dfrac{\pi}{20}$ at $(x,t)=(0,0)$
Thus $Ak=\dfrac{\pi}{20}$
$\implies A=0.025m$
Thus the correct answer is option A.

A harmonic oscillator vibrates with amplitude of 4 cm and performs 150 oscillations in one minute. If the initial phase is $45\circ$ and it starts moving away from the equation of motion is 

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

  1. $\displaystyle 0.04\, sin\, \left ( 5 \pi t\, +\, \frac{\pi}{4} \right )$

  2. $\displaystyle 0.04\, sin\, \left ( 5 \pi t\, -\, \frac{\pi}{4} \right )$

  3. $\displaystyle 0.04\, sin\, \left ( 4 \pi t\, +\, \frac{\pi}{4} \right )$

  4. $\displaystyle 0.04\, sin\, \left ( 4 \pi t\, -\, \frac{\pi}{4} \right )$

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Correct Option: A
Explanation:

Equation of motion of harmonic oscillator is $y=Asin(\omega t+\phi _0)$

It is given that $A=0.04m$
and $\phi _0=54^{\circ}=\dfrac{45}{180}\pi=\dfrac{\pi}{4}$
$\omega=2\pi\nu=2\pi\times \dfrac{150}{60}=5\pi$
Thus $y=0.04sin(5\pi t+\dfrac{\pi}{4})$
Hence the answer is option A.