Tag: waves

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Questions Related to waves

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

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 

  1. $3 cm$
  2. $ 4 cm$
  3. $ 5 cm$
  4. $7 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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, 
$A=\sqrt{A _1^2 +A _2^2}=\sqrt{3^2+4^2}$

$A=5 cms$


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

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

  1. 1

  2. 5/6

  3. 3/4

  4. 6/7
Reveal answer Fill a bubble to check yourself
D Correct answer
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}$