Tag: time period, frequency and amplitude of sound

Quizzes (0) Questions (193)

Questions Related to time period, frequency and amplitude of sound

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

The time period of a system executing wave motion depends upon

  1. elasticity

  2. inertia

  3. Amplitude

  4. restoring force
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. $E \propto n$

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

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

  4. $E \propto \dfrac{1}{log n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. 25 Hz

  2. 50 Hz

  3. 100 Hz

  4. 75 Hz
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

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

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

  1. 10 Sec

  2. 1 Sec

  3. 0.1 Sec

  4. 0.01 sec
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

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.

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

  2. $4.47\ Pa$

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

  4. $5.67\ Pa$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The resultant amplitude of two waves with different frequencies is not a simple sum. However, if interpreting as phasors or peak pressure values, the maximum resultant amplitude is sqrt(P1^2 + P2^2 + 2*P1*P2*cos(phi)). With P1=2e4, P2=4e4, and phase difference, the calculation yields approximately 4.47e4 Pa.

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

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.

  1. 2 sec

  2. 1 sec

  3. 0.5 sec

  4. 1/4 sec
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $1.5 \times 10^3 Hz$

  2. $6 \times 10^8 Hz$

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

  4. $7.5 \times 10^{15} Hz$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$

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

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

  1. f = T

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

  3. T = f$^2$

  4. T = f$^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.  

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

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

  1. is doubled

  2. increases 4 times

  3. decreases 4 times

  4. is halved
Reveal answer Fill a bubble to check yourself
D Correct answer
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.