Tag: waves

Questions Related to waves

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 )

intensity and loudness waves physics

  1. $50 dB$

  2. $76 dB$

  3. $68 dB$

  4. $48 dB$

Show answer
Correct Option: C
Explanation:

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

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

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

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

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

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

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? 

intensity and loudness waves physics

  1. 1000

  2. 100

  3. 10,000

  4. None of these

Show answer
Correct Option: A
Explanation:

The intensity of sound in decibel is given by: 


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

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

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

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

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]

intensity and loudness waves physics

  1. $\displaystyle\frac{9r}{10}$

  2. $\displaystyle\frac{r}{10}$

  3. $\displaystyle\frac{3r}{5}$

  4. $\displaystyle\frac{4r}{5}$

Show answer
Correct Option: A
Explanation:

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

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

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

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

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

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

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

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

  1. True

  2. False

Show answer
Correct Option: B
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 $\sqrt(\mu)$

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

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

  1. $20 N$

  2. $40 N$

  3. $10 N$

  4. $120 N$

Show answer
Correct Option: B
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 $4 \times 10 = 40 N$

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

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

  1. 250.5 and 249.5

  2. 249.5 and 250.5

  3. 124.5 and 125.5

  4. 125.5 and 124.5

Show answer
Correct Option: D

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?

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

  1. 124 : 125

  2. 250 : 313

  3. 5 : 3

  4. 41 : 57

Show answer
Correct Option: A
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 $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

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

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

  1. remains same

  2. increases by 2%

  3. decreases by 2%

  4. frequency becomes imaginary

Show answer
Correct Option: C
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)

The sonometer wire is vibrating in the second overtone. The length of the wire in terms of wavelength is:

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

  1. $L=\lambda/2$

  2. $L=5\lambda/2$

  3. $L=3\lambda/2$

  4. $L=3\lambda$

Show answer
Correct Option: C
Explanation:

second overtone = third harmonic.
Thus, the length of the wire is $L=3\lambda/2$

The correct option is (c)

The frequency of vibration of a sonometer wire is inversely proportional to tension in the wire

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

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

Thus, frequency of vibration is directly proportional to $\sqrt(T)$