Tag: waves

Quizzes (32) Questions (233)

Questions Related to waves

Multiple choice intensity and loudness waves physics

A dog while barking delivers about $1 mW$  of power. If this power is uniformly distributed over a hemispherical area, the sound level at a distance of $5 m$ is (given  10 log$ _{10}$ 6.37 $=$0.8 )

  1. $50 dB$
  2. $76 dB$
  3. $68 dB$
  4. $48 dB$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$I = \dfrac{Power}{Surface Area}$

$I = 6.37 \times 10^{-6} Wm^{-2}$

$I _{o} = 10^{-12} Wm^{-2}$

$SL = 10 log (\dfrac{I}{I _{o}})$

$SL = 10 log (\dfrac{6.37 \times 10^{-6}}{10^{-12}})$

$SL = 10 log (6.37) + 10 log(10^{6})SL = 8 + 60 = 68\ dB$

Multiple choice intensity and loudness waves physics

When a sound wave enters the ear, it sets the eardrum into oscillation, which  in turn causes oscillation of 3 tiny bones in the middle ear called ossicles. This oscillation is finally transmitted to the fluid filled in inner portion of the ear termed as inner ear, the motion of the fluid disturbs hair calls within the inner ear which transmit nerve impulses to the brain with information that a sound is present. The three bones present in the middle ear are named as hammer, anvil and stirrup. Out of these the stirrup is the smallest one and this only connects the middle  ear to inner ear as shown in the figure below. The area of stirrup and its extent of connection with the inner ear limits the sensitivity of the human ear. Consider a person's eat whose moving part of the eardrum has an area of about 43 mm$^{2}$ and the area of stirrup is about 3.2 mm$^{2}$. The mass of ossicles is negligible. As a result, force  exerted by sound wave in air on eardrum and ossicles is same as the force exerted by ossicles on the inner ear. Consider a sound wave having maximum pressure fluctuation of $3\times10^{-2}$ Pa from its normal equilibrium pressure value which is wqual to $10^{5}$ Pa. Frequency of sound wave is 1200 Hz. 
Data: Velocity of sound wave in air is  332 m/s. Velocity of sound wave in fluid (present in inner ear) is 1500 m/s. Bulk modulus of air is $1.42\times10^{5}$ Pa. Bulk modulus of fluid is $2.18\times10^{9}$ Pa. 


If then person is using an hearing aid, which increase the sound intensity level by 30 dB, then by what factor the intensity of given sound wave change as perceived by inner ear? 

  1. 1000

  2. 100

  3. 10,000

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intensity of sound in decibel is given by: 


$I _{dB}=10log(\dfrac{I}{I _0})$

Hence, $30=10log(\dfrac{I}{I _0})$

$\implies I=10^3I _0=1000I _0$

$\implies \dfrac{I}{I _0}=1000$

Multiple choice intensity and loudness waves physics

A bird is singing on a tree and a man is hearing at a distance $'r'$ from the bird. Calculate the displacement of the man towards the bird so that the loudness heard by man increases by $20\;dB$.
[Assume that the motion of man is along the line joining the bird and the man]

  1. $\displaystyle\frac{9r}{10}$
  2. $\displaystyle\frac{r}{10}$
  3. $\displaystyle\frac{3r}{5}$
  4. $\displaystyle\frac{4r}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Loudness $\beta=10\;log _{10}\displaystyle\frac{I}{I _0}$

$\therefore\;\beta _2-\beta _1=10\;log _{10}\displaystyle\frac{I _2}{I _1}$

$\because\;I=\displaystyle\frac{P}{4\pi r^2}\;\therefore\;\displaystyle\frac{I _2}{I _1}=\displaystyle\frac{r _1^2}{r _2^2}$

$\therefore\;(\beta+20)-\beta=10\;log _{10}\displaystyle\frac{r^2}{r _2^2}=20\;log _{10}\displaystyle\frac{r}{r _2}$

$\Rightarrow \displaystyle\frac{r}{r _2}=10\Rightarrow r _2=0.1\;r$

$\therefore\;shift=r-0.1\;r=0.9\;r$.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The frequency of vibration of a sonometer wire is directly proportional to linear density of the wire:

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt{(T/\mu)}; \mu $ is the linear density of the material of the wire
 
Thus, frequency of vibration is inversely proportional to $\sqrt(\mu)$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The tension in a piano wire is $10 N$. The tension in a piano wire to produce a node of double frequency is

  1. $20 N$
  2. $40 N$
  3. $10 N$
  4. $120 N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For frequency of oscillation of wire $n \propto \sqrt{T}$. 

Here $T$ is tension in the wire. In order to increase frequency twice, tension needs to be made $4$ times. 
So, new tension must be $4 \times 10 = 40 N$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A knife edge divides a sonometer wire in two parts which differ in length by 2 mm. The whole length of the wire is 1 meter. The two parts of the string when sounded together produce one beat per second. Then the frequency of the smaller and longer pans.in Hz,are

  1. 250.5 and 249.5

  2. 249.5 and 250.5

  3. 124.5 and 125.5

  4. 125.5 and 124.5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the frequency formula f = (1/2L) * sqrt(T/m), the difference in frequencies for two segments of lengths L1 and L2 is 1 Hz. Given L1 + L2 = 1m and L1 - L2 = 0.002m, we find L1 = 0.501m and L2 = 0.499m. The frequencies are proportional to 1/L, leading to 124.5 Hz and 125.5 Hz.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire of length $l _1$ vibrates with a frequency 250 Hz. If the length of wire is increased then 2 beats/s are heard. What is ratio of the lengths of the wire?

  1. 124 : 125

  2. 250 : 313

  3. 5 : 3

  4. 41 : 57
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The frequency of sonometer wire is given by
$n=\dfrac{p}{2l}\sqrt{(\dfrac{T}{m})}$
or $n\propto \dfrac{1}{l}$     ...(i)
$\therefore \dfrac{n _1}{n _2}=\dfrac{l _2}{l _1}$
or $\dfrac{250}{250-2}=\dfrac{l _2}{l _1}$
or $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The tension in the sonometer wire is decreased by 4% by loosening the screws. It fundamental frequency

  1. remains same

  2. increases by 2%

  3. decreases by 2%

  4. frequency becomes imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt(T/\mu)$

The fractional change in frequency with change in tension is given by $\Delta f/f = -(\Delta T/2T)$.

Thus, the frequency decreases by 2%

The correct option is (c)