a) suppose a hearing aid amplifies the amplitude of a sound outside the ear by a

Question

a) suppose a hearing aid amplifies the amplitude of a sound outside the ear by a factor of twelve, what would be the sound pressure inside the ear given an external sound level of 60 dB SPL?

b) Volker normally has a threshold of hearing for 1 KHz sinusoid of 30 uPa, after visiting a disco with loud music, Volker's threshold of hearing for a 1 KHz tone is temporarily raised numerically by 15 dB. what is his new threshold of hearing in uPa?

c) the normal threshold of hearing at 200 KHzis 63.3 uPa. Stuart has a threshold that is numerically 4 dB better than average at this frequency. what sound pressure (in pa or uPa) is the least intense that Stuart can hear at 200 Hz?

a) Suppose a hearing aid amplifies the amplitude of a sound outside the ear by a factor of twelve, what would be the sound pressure inside the ear given an external sound level of 60 dB SPL? b) Volker normally has a threshold of hearing for 1 KHz sinusoid of 30 mu Pa, after visiting a disco with loud music, Volker's threshold of hearing for a 1 KHz tone is temporarily raised numerically by 15 dB. what is his new threshold of hearing in mu Pa? c) the normal threshold of hearing at 200 KHz is 63.3 uPa. Stuart has a threshold that is numerically 4 dB better than average at this frequency. what sound pressure (in pa or mu Pa) is the least intense that Stuart can hear at 200 Hz?

Explanation / Answer

a) Given a hearing aid amplifies the amplitude of a sound outside the ear by a factor of twelve.

Also given

An external sound level of 60 dB SPL

We know that by doubling the sound pressure raises the sound pressure level by 6 dB

Hence by doing tweleve times the sound presusre raises the sound pressure level by 36 dB

Therefore the sound pressure inside the ear given an external sound level of 60 dB SPL = 60 + 36 = 96 dB

b) Given

Volker normally has a threshold of hearing for 1 KHz sinusoid of 30 uPa, after visiting a disco with loud music, Volker's threshold of hearing for a 1 KHz tone is temporarily raised numerically by 15 dB.

Let his new threshold of hearing in uPa= x

We know that

1 Pa = 94 dB

And Therefore 1dB = 1/94Pa = 1000000/94 uPa

So 15 dB = 1000000/94 * 15 = 16.4835 * 10^4 uPa.

Hence his new threshold of hearing in uPa= x = 16.4835 * 10^4 uPa.

c) Given

The normal threshold of hearing at 200 KHzis 63.3 uPa. Stuart has a threshold that is numerically 4 dB better than average at this frequency

Sound pressure (in pa or uPa) is the least intense that Stuart can hear at 200 Hz = 63.3 uPa + 4 dB

= 63.3 uPa + 2659 .57

= 2722.87 uPA

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