Tag: beats in sound waves

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Multiple choice physics beats in sound waves

The musical interval between tube two tine of frequency $400\ Hz$ and $200\ Hz$ is:

  1. $2$
  2. $200$
  3. $1$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The musical interval is defined as the ratio of the two frequencies: 400 Hz / 200 Hz = 2.

Multiple choice physics beats in sound waves

A key of mechanical piano is first struck gently and then struck again but much harder this time. In the second case :

  1. sound will be louder but pitch will not be different

  2. sound will be louder and the pitch will also be higher

  3. sound will be louder but pitch will be lower

  4. both loudness and pitch will remain unaffected
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Loudness or softness of a sound is determined basically by its amplitude. The amplitude of the sound wave depends upon the force with which an object is made to vibrate. If we strike a table lightly, we hear a soft sound because we produce a sound wave of less energy (amplitude). Pitch depends upon frequency received by the human ear.

Multiple choice physics beats in sound waves

The characteristics of sound with the help of which we can distinguish between a SHRILL note and a GRAVE note is

  1. Loudness

  2. Pitch

  3. Quality

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

Pitch is the characteristic of sound that allows us to distinguish between a shrill note (high frequency) and a grave note (low frequency).

Multiple choice physics beats in sound waves

How Jal Tarang produces sound?

  1. In Jal-tarang musical instrument the cup containing minimum water produces the sound of lowest frequency.

  2. In Jal-tarang musical instrument the cup containing minimum water produces the sound of lowest pitch.

  3. In Jal-tarang musical instrument the cup containing minimum water produces the sound of lowest frequency. As the amount of water in the cup goes on increasing, the frequency of the sound produced also goes on increasing.

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

n Jal-tarang musical instrument the cup containing minimum water produces the sound of lowest frequency or lowest pitch. As the amount of water in the cup goes on increasing, the frequency (or the pitch) of the sound
produced also goes on increasing.

Multiple choice physics beats in sound waves

Beats are heard at the rate of $12$ every $5$ seconds when two open organ pipes of lengths $84\ cm$ and $85\ cm$ are sounded together in their fundamental modes. Find the velocity of sound in air. (An open organ pipe has an antinode at each end.

  1. $342.7\ m/s$.
  2. $300.7\ m/s$.
  3. $350.7\ m/s$.
  4. $400.7\ m/s$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an open organ pipe, fundamental frequency f = v / 2L. f1 = v / (2 * 0.84), f2 = v / (2 * 0.85). Beat frequency |f1 - f2| = (v/2) * (1/0.84 - 1/0.85) = 12/5 = 2.4. (v/2) * (0.01 / (0.84 * 0.85)) = 2.4. v = 2.4 * 2 * 0.84 * 0.85 / 0.01 = 342.72 m/s.

Multiple choice physics beats in sound waves

An under water swimmer sends a sound signal to the surface. It is produces 5 betas/sec when compared with fundamental tone of a pipe of $20$ cm length closed at one end what is wavelength of sound in water. (take V water=$1500 m/sec$ Vair=$360 m/sec$)

  1. $3.3 m$ or $3.37 m$
  2. $4.4 m$ or $4.47 m $
  3. $2.5 m$ or $2.7 m$
  4. $1 m$ or $1.7 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A pipe of length 0.2 m closed at one end has fundamental frequency f = v_air / (4L) = 360 / (4 * 0.2) = 450 Hz. The swimmer's signal frequency f_s satisfies |f_s - 450| = 5, so f_s = 455 or 445 Hz. Wavelength in water lambda = v_water / f_s = 1500 / 455 = 3.296 m or 1500 / 445 = 3.37 m.

Multiple choice physics beats in sound waves

identify the part which vibrates to produce sound in the following instruments.

  1. tabla or dholak

  2. flute

  3. tuning fork

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

A tuning fork produces sound through the vibration of its metal prongs. While the other instruments listed also produce sound through vibration, the tuning fork is the classic example of a simple vibrating object used in physics.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two sound sources are moving in opposite direction with velocity $v _1$ and $v _2$ $(v _1>v _2)$. Both are moving away from a stationary observer.the frequency of both the source is $900\ Hz$. What is the value of $v _1 - v _2 $  so that the beat frequency observed will be $6\ Hz$  ?


Speed of sound =$300\ ms^{-1}$

  1. $1\ ms^{-1}$
  2. $4\ ms^{-1}$
  3. $3\ ms^{-1}$
  4. $2\ ms^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$f _1 = 900(\dfrac{300}{300+v _1})$

$\implies f _1= 900(1+ \dfrac{v _1}{300})^{-1}$
$\implies f _1 = 900 - 3v _1$
Similarly 
$f _2 = 900 - 3v _2$
So,
$f _1 -f _2 = 6$
$3(v _1 - v _2)= 6$
$\implies v _1 - v _2 = 2\ ms^{-1}$

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

A sources of sonic oscillations with frequency n= $1700$ Hz and a receiver are located on the same normal to a wall. Both the source and receiver are stationary, and the wall recedes from the source with velocity u= $6.0$ cm/s. Find the beat frequency registred by the receiver. The velocity of sound is equal to $v= 340$ m/s.

  1. $0.2$ Hz
  2. $0.3$ Hz
  3. $0.4$ Hz
  4. $0.6$ Hz
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The wall acts as a moving reflector. The frequency of the sound reflected by the wall is f' = f * (v + u) / (v - u). The beat frequency is the difference between the reflected frequency and the source frequency, which simplifies to f_beat = f * (2u / (v - u)). Plugging in f=1700, u=0.06 m/s, and v=340 m/s gives 1700 * (0.12 / 339.94), which is approximately 0.6 Hz.