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Allushta [10]
3 years ago
9

While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at th

at point. The lowest-frequency standing wave that could be present on the string vibrates at Group of answer choices a. twice the fundamental frequency. b. the fundamental frequency. c. three times the fundamental frequency. d. four times the fundamental frequency. e. There is not enough information to decide.
Physics
1 answer:
fiasKO [112]3 years ago
4 0

Answer:

Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.

Explanation:

Before touching the midpoint of the string, the string vibrates with one loop.

Fundamental frequency, f1 = v/(2*L)

Now, when the midpoint of the guitar string was touched, the string vibrates with two loops.

Hence, f2 = 2*v/(2*L)

f2 = 2*f1

Therefore, compared to the fundamental frequency the frequency would be double.

Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.

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Explanation:

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For Destructive interference

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where \lambda =wavelength

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for second order maxima i.e. m=2

For smallest separation taking \lambda =400 nm, \theta =90^{\circ}

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Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

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