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

  1. $Zero$

  2. $80\, m/s$

  3. $320\, m/s$

  4. $160\, m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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 

  1. 20 cm

  2. 25 cm

  3. 10 cm

  4. 15 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The centripetal force is provided by the spring force: m * omega^2 * (r + x) = k * x, where r is the original length and x is the elongation. For omega, x1 = 1 cm. For 2 * omega, x2 = 5 cm. Thus, m * omega^2 * (r + 1) = k * 1 and m * (2 * omega)^2 * (r + 5) = k * 5. Dividing the equations: 4 * (r + 5) / (r + 1) = 5. Solving for r: 4r + 20 = 5r + 5, so r = 15 cm.

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

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).

  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. $y\, =\, 0.5\, cos\, (100 \pi t\, -\, 10 \pi x)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

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 

  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. $\displaystyle 0.04\, sin\, \left ( 4 \pi t\, -\, \frac{\pi}{4} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

Which of the following statements is correct?

  1. Longitudinal waves consist of crests and troughs

  2. In case of transverse waves, the particles of the medium vibrate at right angles to the direction of wave

  3. Transverse waves are produced when a tuning fork is struck in air

  4. Longitudinal waves are produced when a stone is dropped on the surface of water in a pond
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Answer is C.

A transverse wave is a wave in which the medium vibrates at right angles to the direction that the wave travels. An example of a transverse wave is a wave in a rope held with a hand on one end and tied to a pole on another end.
In this wave, energy is provided by a persons hand moving one end of the rope up and down. The direction of the wave is down the length of the rope away from the persons hand. The rope itself moves up and down as the wave passes through it.
The characteristic described in statement c is a property of all transverse waves, but not necessarily of all mechanical waves. A mechanical wave can also be longitudinal.
Hence, option C is correct and rest of the statements are incorrect.

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

Common balance reading on earth of an object is 'X' and the spring balance reading is 'Y'. The set up is transferred to moon. What reading does the common balance and spring balance give?

  1. <span>$\dfrac{X}{6}, Y$</span>

  2. <span>$X, \dfrac{Y}{2}$</span>

  3. <span>$X, \dfrac{Y}{6}$</span>

  4. <span>$X, \dfrac{Y}{3}$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A common balance (beam balance) measures mass, which is invariant under gravity. A spring balance measures weight (W = mg), which depends on the local acceleration due to gravity g. On the moon, g is 1/6 of that on Earth, so the spring balance reading becomes Y/6.

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

Two vibrating strings of the same material but length $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension. Both the string vibrate in their fundamental modes, the one of length $L$ with frequency ${v} _{1}$ and other with frequency ${v} _{2}$. The ratio ${v} _{1}/{v} _{2}$ is given by

  1. $2$

  2. $4$

  3. $8$

  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For vibrating string 
frequency $\alpha \dfrac{1}{lenght} \sqrt{\dfrac{Tension}{mass\ per\ unit\ lenght}}$ ---- $(1)$
Since area of first wire is $\dfrac{\pi (2)^{2}}{\pi 1^{2}} = 4$ times pf second wire
its mass per unit length is also $4$ times 
i.e. $\mu _{1}= 4 \mu$ $\mu _{2}= \mu$
$v _{1} = \dfrac{1}{L} \sqrt{\dfrac{T}{4\mu}}= \dfrac{1}{\mu} \sqrt{\dfrac{T}{\mu}}$
$v _{2}= \dfrac{1}{2} \sqrt{\dfrac{T}{\mu}}$
$\dfrac{v _{1}}{v _{2}}= \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}} \times  \dfrac{2l}{1} \sqrt{\dfrac{\mu}{T}}$
$=1$
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A sonometre wire resonates with a given tuning forck forming standing waves with five antinodes between the two bridges when a mass of $9kg$is suspended from the wire. When this mass is replaced by mass $M$, the wire resonates with the same positions of the bridges. Then find the value of square roof of $M$.

  1. $5$

  2. $10$

  3. $25$

  4. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The frequency of vibration of a string 

$n=\dfrac{p}{2l}\sqrt{\dfrac{T}{m}}$
Also number of loops = Number of antinodes.
Hence with 5 antinodes and hanging mass of 9 kg. we have p=5 and T=9g
So,
$n _1=\dfrac{5}{2l}\sqrt{\dfrac{9g}{m}}$
With 3 antinodes and hanging mass M we have p=3 and T=Mg so,
$n _2=\dfrac{3}{2l}\sqrt{\dfrac{Mg}{m}}$
$\because n _1=n _2$
$\dfrac{5}{2l}\sqrt{\dfrac{9g}{m}}=\dfrac{3}{2l}\sqrt{\dfrac{Mg}{m}}$
Squaring both side we get
$25\times9=9\times M$
$M=25\ kg$