Tag: time period, frequency and amplitude of sound

Questions Related to time period, frequency and amplitude of sound

The time period of a system executing wave motion depends upon

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. elasticity

  2. inertia

  3. Amplitude

  4. restoring force

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Correct Option: A
Explanation:

Time period of oscillations is directly proportional to square root of modulus of elasticity and inversely proportional to density. Thus option (a) is the correct option

The total energy of sound waves E is related to their frequency n as

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. $E \propto n$

  2. $E \propto \dfrac{1}{n}$

  3. $E \propto \dfrac{1}{n^2}$

  4. $E \propto \dfrac{1}{log n}$

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Correct Option: A
Explanation:

Energy of a wave is related to its frequency using the formula E = hf, f is the frequency of the wave

A particle performs 20 vibrations in 5 secs. The time period of vibration is

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. 2.5 s

  2. 0.1 s

  3. 0.25 s

  4. 4 s

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Correct Option: C
Explanation:

Time taken for 1 vibration = 5/20 = 0.25 s

The correct option is (c)

A transverse wave of frequency 50 Hz is reflected from a wall. 50% of the energy of the wave is lost at the wall. The frequency of the reflected wave will be

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. 25 Hz

  2. 50 Hz

  3. 100 Hz

  4. 75 Hz

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Correct Option: B
Explanation:

Loss in energy only implies loss in amplitude and not in frequency. Since frequency is a characteristic of the source

The correct option is (b)

A particle vibrates 10 times in 1 sec. What is the time period of vibration of the particle

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. 10 Sec

  2. 1 Sec

  3. 0.1 Sec

  4. 0.01 sec

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Correct Option: C
Explanation:

Time period of vibration is time taken for 1 vibration

Thus 10 vibrations are completed in 1 sec or 1 vibration will taken only 1/10=0.1 sec

The correct option is (c)

Two sound waves having pressure
$P _{1}=2 \times 10^{4} \sin (2\pi \times 10^{4}\ t)Pa$ and 
$P _{2}=4 \times 10^{4} \sin (3\pi \times 10^{4}\ t+\pi /6)Pa$
superimpose with each other. Find the amplitude of resultant wave.

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. $4.47\times 10^{4}\ Pa$

  2. $4.47\ Pa$

  3. $5.67\times 10^{4}\ Pa$

  4. $5.67\ Pa$

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Correct Option: A

Two tuning forks of frequency 256 and 258 vibrating per second are sounded together, then time interval between consecutive maxima heard by the observer is.

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. 2 sec

  2. 1 sec

  3. 0.5 sec

  4. 1/4 sec

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Correct Option: C
Explanation:

$\begin{array}{l} { n _{ 1 } }=256 \ { n _{ 2 } }=258 \ Both\, tuning\, forks\, are\, making\, beats \ so,\, we\, know\, that \ { t _{ \left( { for\, \, \max  . } \right)  } }=\dfrac { 1 }{ { { n _{ 2 } }-{ n _{ 1 } } } } \sec   \ =\dfrac { 1 }{ { 258-256 } }  \ =0.5\, \sec   \end{array}$

Hence, the option $C$ is the correct answer.

The frequency of light of wave length $5000{A^\circ}$ is :

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. $1.5 \times 10^3 Hz$

  2. $6 \times 10^8 Hz$

  3. $6 \times 10^{14} Hz$

  4. $7.5 \times 10^{15} Hz$

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Correct Option: C
Explanation:

We know that $ n = \dfrac{v}{\lambda } $
where, $ v = $ velocity of light
$\lambda  =$ wavelenght
So, $ n = \dfrac{ 3 \times  10^{8}}{5000 \times  10^{-10} } = 6 \times 10^{14} Hz$

The relation between frequency(f) and time period(T) is given by

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. f = T

  2. f = $\frac{1}{T}$

  3. T = f$^2$

  4. T = f$^3$

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Correct Option: B
Explanation:

The frequency (f) of a wave is the number of full wave forms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.  
Time Period (T) is the number of seconds per waveform, or the number of seconds per oscillation.  It is clear that frequency and time period are reciprocals. 
That is, T=1/f.  

If the frequency of sound wave in air is doubled, its wavelength.

physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

  1. is doubled

  2. increases 4 times

  3. decreases 4 times

  4. is halved

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Correct Option: D
Explanation:

Answer is D.

A sound wave has a speed that is mathematically related to the frequency and the wavelength of the wave. The mathematical relationship between speed, frequency and wavelength is given by the following equation.
Speed = Wavelength * Frequency
That is, Frequency = speed / Wavelength. The frequency and wavelength are inversely proportional to each other.
So, when the frequency is doubled, the wavelength will be halved.