Tag: waves

Questions Related to waves

The displacement of an elastic wave is given by the function $y= 3\ sin \omega t+4\ cos\omega t$, where $y$ is in $cm$ and $t$ is in $s$. The resultant amplitude is 

problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

  1. $3 cm$

  2. $ 4 cm$

  3. $ 5 cm$

  4. $7 cm$

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Correct Option: C
Explanation:

Here, we have given: $y=3sin\omega t+4cos\omega t$


So, two components are there, $3sin\omega t $ and $4cos\omega t$


where, individual amplitudes are given by
$A _1= 3 cms$ and $A _2=4 cms .$

so , resultant amplitude will be, 
$A=\sqrt{A _1^2 +A _2^2}=\sqrt{3^2+4^2}$

$A=5 cms$


A string of length $l$ is fixed at both ends and is vibrating in second harmonic. The amplitude at antinode is $2\ mm$. The amplitude of a particle at distance $l/8$ from the fixed end is :

problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

  1. $5\sqrt 2\ mm$

  2. $\dfrac{5}{\sqrt 2}\ mm$

  3. $5\ mm$

  4. $\dfrac{10}{\sqrt 2}\ mm$

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Correct Option: B

Equations of a stationary wave and a travelling wave are $y _1=1\,sin(kx)\,cos (\omega t)$ and $y _2=a\,sin\,(\omega t-kx)$.The phase difference between two points $x _1=\dfrac{\pi}{3k}$ and $x _2=\dfrac{3 \pi}{2k}$ is $\phi _1$ for the first wave and $\phi _2$ for the second wave.The ratio $\dfrac{\phi _1}{\phi _2}$ is

problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

  1. 1

  2. 5/6

  3. 3/4

  4. 6/7

Show answer
Correct Option: D
Explanation:

Phase difference between two points in a standing wave =$n\pi$

Where n is number of nodes between two points.
Given points are $x _1 = \cfrac{\pi}{3k} = \cfrac{60}{k}$
$x _2 = \cfrac{3\pi}{2k} = \cfrac{210}{k}$
Equation of the standing wave
$ y _1 = a \sin kx \cos \omega t$
At node points $ kx =n\pi$
$ x = \cfrac{n\pi}{k} \quad (n=0,1,2,3...)$
So nodes are =$ \cfrac{\pi}{k} , \cfrac{2\pi}{k} ....$
$ =  \cfrac{180}{k} , \cfrac{360}{k} ....$
Since there is only one node between phase difference  $ \phi _1 = \pi$
For travelling wave $ \phi _2  = \cfrac{2\pi}{\lambda} \triangle x$
From the equation 
$y _2 = a \sin (\omega t - kx)$
$ k = \cfrac{2\pi}{\lambda}$
$ \therefore \phi _2 = k[x _2 - x _1] = k[\cfrac{3\pi}{2k} - \cfrac{\pi}{3k}] = \cfrac{7}{6}\pi$
$ \therefore \cfrac{\phi _1}{\phi _2} = \cfrac{\pi}{\cfrac{7}{6}\pi} = \cfrac{6}{7}$