Question

A 60 year old person has a threshold of hearing of 95.0 dB for a sound with frequency f=10,000 Hz. By what factor must the intensity of a sound wave of that frequency, audible to a typical young adult, (sound level=43.0 dB) be increased so that it is heard by the older person.
If you know the sound level in dB, then you can compute the sound intensity and take the ratio.


Helpful Hint*
In the problem, they gave the intensity as decibels but want to know by what factor to increase the intensity (I) for the young person to get a sound that the older person can hear.

Ignore f completely.

Decibels (dB) need to be converted to I using this equation:
I=(10^(dB/10))*10^-12
Do this for each dB and divide the older person's dB by the younger's.

Answer 1

Answer:

The intensity increased by a factor of 158489

Explanation:

Given that,

Sound level = 95.0 dB

Sound level = 43.0 dB

Frequency = 10000 Hz

We need to calculate the ratio of sound intensity

Using formula of sound level

$sound\ level =10 log\dfrac{I}{I_{0}}$

Put the value into the formula

$95.0=10\ log\dfrac{I_{1}}{I_{0}}$...(I)

$43.0=10\ log\dfrac{I_{2}}{I_{0}}$.....(II)

Subtracting these equations

$52.0=10\ log\dfrac{I_{1}}{I_{2}}$

$\log\dfrac{I_{1}}{I_{2}}=5.2$

Taking inverse log

$\dfrac{I_{1}}{I_{2}}=10^{5.2}$

$\dfrac{I_{1}}{I_{2}}=158489$

Hence, The intensity increased by a factor of 158489