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artcher [175]
3 years ago
6

A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pi

tch. The first-chair player's note is exactly 440 Hz, and 2.10 beats per second are heard.What are the two possible frequencies of the other players note?
Physics
1 answer:
KIM [24]3 years ago
4 0

To develop this problem, it is necessary to apply the concepts related to Beat

The Beat is an acoustic phenomenon that is generated by two sine waves interfering with slightly different frequencies. The beat frequency is equal to the difference in the frequencies of the two original waves:

f_{bat} = |(f_1 \pm f_B)|

Our values are given as

f_1 = 440Hz

f_B = 2.1Hz

For the particular case we have two possible frequencies:

1) f_{bat} = |(f_1 - f_B)|

f_{bat} = |(440 - 2.1)|

f_{bat} = 437.9Hz

2) f_{bat} = |(f_B + f_1)|

f_{bat} = |(440 +2.1)|

f_{bat} = 442.1Hz

Therefore the two possibles frequencies of the other players note are 437.9Hz and 442.1Hz

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Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

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Moment after shock

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   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

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   v_{f}= 4.4 600 / (600 + 500)

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With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

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After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

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