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wolverine [178]
3 years ago
15

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, th

e note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 602 N to 564.00 N, what beat frequency is heard when the hammer strikes the two strings simultaneously? beats/s
Physics
1 answer:
Georgia [21]3 years ago
8 0

Explanation:

Given that,

Frequency in the string, f = 110 Hz

Tension, T = 602 N

Tension, T' = 564 N

We know that frequency in a string is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{m/L}}, T is the tension in the string

i.e.

f\propto\sqrt{T}

\dfrac{f}{f'}=\sqrt{\dfrac{T}{T'}}, f' is the another frequency

{f'}=f\times \sqrt{\dfrac{T'}{T}}

{f'}=110\times \sqrt{\dfrac{564}{602}}

f' =106.47 Hz

We need to find the beat frequency when the hammer strikes the two strings simultaneously. The difference in frequency is called its beat frequency as :

f_b=|f-f'|

f_b=|110-106.47|

f_b=3.53\ beats/s

So, the beat frequency when the hammer strikes the two strings simultaneously is 3.53 beats per second.

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zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

|

B1 at 20km/h

|

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

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