Tag: reflection of waves

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Questions Related to reflection of waves

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

A set of 3 standing waves 5, 10 and 15 Hz are to be setup on a string fixed at one end. One of these frequencies are suppressed, while passing through it. Identify them:

  1. 5 Hz

  2. 10 Hz

  3. 15 Hz

  4. All the frequencies will pass through them
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a string fixed at one end, only odd harmonics are allowed and even harmonics are suppressed.

Thus the 10Hz standing wave is suppressed,
The correct option is (b)

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

Find the number of beats produced per sec by the vibrations $x _1=A\sin (320\pi t)$ and $x _2=A\sin (326\pi t)$.

  1. 3

  2. 4

  3. 5

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$X _1=Asin(320\Pi t)$

$X _2=Asin(326\Pi t)$
On comparing it with general equation.
$X=Asin(wt)$
Then, $w _1=320\Pi $
$w _1=2\Pi f$
frequency=160 Hz
Similarly,
$w _2=326\Pi $
$w _2=2\Pi f$
frequency=163 Hz
No of beats=163-160=3

Multiple choice physics oscillations and waves standing waves in strings standing waves reflection of waves

In an organ pipe(may be closed or open) of $99$ cm length standing wave is setup, whose equation is given by longitudinal displacement.
$\xi =(0.1mm)\cos \dfrac{2\pi}{0.8}(y+1cm)\cos 2\pi (400)t$
where y is measured from the top of the tube in meters and t is second. Here $1$cm is the end correction.
The air column is vibrating in :

  1. First overtone

  2. Fifth harmonic

  3. Third harmonic

  4. Fundamental mode
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is of the form cos(k(y+c))cos(wt). The angular frequency w = 2*pi*400, so f = 400 Hz. The wave number k = 2*pi/0.8, so wavelength lambda = 0.8 m = 80 cm. With end correction, the effective length is 99 + 1 = 100 cm. For a closed pipe, L = (2n-1)lambda/4. 100 = (2n-1)80/4 = (2n-1)20. 5 = 2n-1, so n=3. This corresponds to the 5th harmonic.