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

A man generates a symmetrical plus in a string by moving his hand up and down. At $t=0$ the point in his hand moves downward. The pulse travels with speed $3 m/s$ on the string & his hands passes $6$ times in eacgh seconds from the mean position. Then the point on the string at a distance $3m$ will reach  its upper extreme first time at time $t=$

  1. $1.25 sec.$

  2. $1 sec.$

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

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

Firstly, we have to draw the pulse wave (see attached diagram).

If hand passes 6 times from the mean position in one second, then we know that string creates 3 wave lengths (λ) or 3 cycles after 1 second.

That means frequency (f) of the wave is $3 Hz.$

Now we can use below equation to calculate the value of wave length.

$V = f\lambda$ (V = velocity of the wave)

$\lambda =\dfrac{ V}{f}$

  $= \dfrac{(3m/s)}{3} = 1 m$

If $\lambda = 1m$, the point which have 3m distance is located at no.(6) (in the diagram).

to reach its upper extreme ----> have to travel 3λ/4 distance

time to travel $3\lambda = 1$ second

time to travel $\lambda = \dfrac{1}{3}$ seconds

time to travel $\dfrac{3\lambda}{4} = (\dfrac{1}{3}) \times (\dfrac{3}{4})$ seconds

 $= \dfrac{1}{4}$ seconds $= 0.25$ seconds

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

A wave moving with constant speed on a uniform string passes the point $x = 0$ with amplitude $\displaystyle A _{0}$, angular frequency $\displaystyle \omega _{0}$ and average rate of energy transfer $\displaystyle P _{0}$. As the wave travels down the string it gradually loses energy and at the point x = $\displaystyle l $, the average rate of energy transfer becomes $\displaystyle \dfrac{P _{0}}{2}$. At the point x = $\displaystyle l$, angular frequency and amplitude are respectively

  1. $\displaystyle \omega _{0}$ and $A _{0}/\sqrt{2}$

  2. $\displaystyle \omega _{0}/\sqrt{2}$ and $A _{0}$

  3. less than $\displaystyle \omega _{0}$ and $A _{0}$

  4. $\displaystyle \omega _{0}/\sqrt{2}$ and $ A _{0}/\sqrt{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The average power of a wave is proportional to the square of the amplitude and the square of the frequency. If the frequency remains constant as the wave travels, the reduction in power must be due to a reduction in amplitude.

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

A stationary wave $y=0.4\sin \cfrac{2\pi}{40}x\cos 100\pi t$ is produced in a rod fixed at both end. The minimum possible length of the rod is given by:

  1. 10 m

  2. $20\sqrt2m$

  3. 20 m

  4. 28 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a standing wave y = A sin(kx) cos(omega*t), the nodes occur where sin(kx) = 0. For a rod fixed at both ends, the length L must be an integer multiple of half-wavelengths. Given k = 2*pi / 40, lambda = 40. Minimum length is lambda / 2 = 20.

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

Two strings A and B with $\mu= 2 \ kg/m$ and $\mu= 8 \ kg/m$ respectively are joined in series and kept on a horizontal table with both the ends fixed. The tension in the string is 200 N. If a pulse of  amplitude 1 cm travels in A towards the junction, then find the amplitude of reflected and transmitted pulse. 

  1. $A _r=2 A _T=7$

  2. $A _r=\dfrac{-1}{3} A _T=\dfrac{2}{3}$

  3. $A _r=8 A _T=9$

  4. $A _r=3 A _t=4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Velocity of wave in string A, ${v _A} = \sqrt {\dfrac{T}{{{\mu _A}}}}  = \sqrt {\frac{{200}}{2}}  = 10\,\,m/s$

Velocity of wave in string B,${v _B} = \sqrt {\dfrac{T}{{{\mu _B}}}}  = \sqrt {\frac{{200}}{8}}  = 5\,\,m/s$
Using $k = \dfrac{w}{v} \Rightarrow {k _A} = 0.1w\,\,and\,{k _B} = 0.2w$
Amplitude of reflected pulse, ${A _B} = \dfrac{{{k _A} - {k _B}}}{{{k _A} + {k _B}}}A = \dfrac{{0.1 - 0.2}}{{0.1 + 0.2}} \times 1 =  - \dfrac{1}{3}$
Amplitude of transmitted pulse,${A _T} = A - \left| {{A _R}} \right| = 1 - \dfrac{1}{3} = \dfrac{2}{3}\,\,cm$

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

A wave travels on a light string. The equation of the wave is Y = A sin(Kx - $\omega$t + 30$^o$). It is reflected from a heavy string tied to an end of the light string at x = 0. If 64% of the incident energy is reflected the equation of the reflected wave

  1. $Y = 0.8 A sin(Kx - \omega \ t + 30^o + 180^o)$

  2. $Y = 0.8 A sin(Kx + \omega \ t + 30^o + 180^o)$

  3. $Y = 0.8 A sin(Kx + \omega \ t - 30^o)$

  4. $Y = 0.8 A sin(Kx + \omega$t + 30^o)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a wave reflects from a denser medium, it undergoes a phase change of 180 degrees. The amplitude of the reflected wave is determined by the energy reflection coefficient (R = sqrt(0.64) = 0.8).

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

What should one do if he wishes to increase the pitch of a string type instrument.
$1$. Increase the length of the string used
$2$. Decrease the gauge of the string used
$3$. Loosen the string
$4$. Tighten the string

  1. $1$ and $4$

  2. $2$ and $4$

  3. $2, 1$ and $4$

  4. $3$ and $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Thicker, tighter strings, have a more "focussed" sound. They reach their resonant frequency more quickly, because the extra tension leaves them less scope to flap around.

Thicker, tighter strings, plucked the same distance, are louder, because they contain more energy. There is more kinetic energy to be transmitted to the sounding board.


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

A stretched string is vibrating at $500$ hertz. If the tension is increased four times, the frequency shall become.

  1. $1,000$ hertz

  2. $500$ hertz

  3. $250$ hertz

  4. $1,500$ hertz
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As frequency $(f)$ is directly proportional to the square root of tension$(t)$.

$f\propto \sqrt t$
$\cfrac{f _1}{f _2}=\sqrt{\cfrac{t _1}{t _2}}$
$\cfrac{500}{f _2}=\sqrt{\cfrac{1}{4}}$
$\implies f=1,000$ hertz