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2 years ago

What beat frequencies are possible with tuning forks of frequencies 255, 258, and 260 hz ?

Physics

1 answer:

zavuch27 [327]2 years ago

7 0

Possible beat frequencies with tuning forks of frequencies 255, 258, and 260 Hz are 2, 3 and 5 Hz respectively.

The beat frequency refers to the rate at which the volume is heard to be oscillating from high to low volume. For example, if two complete cycles of high and low volumes are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 256 Hz and 254 Hz are played simultaneously, a beat frequency of 2 Hz will be detected. A common physics demonstration involves producing beats using two tuning forks with very similar frequencies. If a tine on one of two identical tuning forks is wrapped with a rubber band, then that tuning forks frequency will be lowered. If both tuning forks are vibrated together, then they produce sounds with slightly different frequencies. These sounds will interfere to produce detectable beats. The human ear is capable of detecting beats with frequencies of 7 Hz and below.

A piano tuner frequently utilizes the phenomenon of beats to tune a piano string. She will pluck the string and tap a tuning fork at the same time. If the two sound sources - the piano string and the tuning fork - produce detectable beats then their frequencies are not identical. She will then adjust the tension of the piano string and repeat the process until the beats can no longer be heard. As the piano string becomes more in tune with the tuning fork, the beat frequency will be reduced and approach 0 Hz. When beats are no longer heard, the piano string is tuned to the tuning fork; that is, they play the same frequency. The process allows a piano tuner to match the strings' frequency to the frequency of a standardized set of tuning forks.

Learn more about beat frequency here : brainly.com/question/14157895

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Answer:

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Explanation:

Given that,

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Using formula of force

$F = ma$

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Put the value into the formula

$m=\dfrac{80.5}{1.28}$

$m=62.8\ kg$

(b). If the worker stops pushing at the end of 4.50 s,

We need to calculate the velocity

Using equation of motion

$v =u+at$

Put the value into the formula

$v=0+1.28\times4.50$

$v=5.76\ m/s$

We need to calculate the distance

Using formula of distance

$v = \dfrac{d}{t}$

$d=v\times t$

Put the value into the formula

$d=5.76\times4.20$

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(b). The distance is 24.192 m.

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Answer:

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Answer:

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