A 500 500500- kg kgstart text, k, g, end text object is accelerating to the right at 10 cm / s 2 10 cm/s 2 10, start text, space
2 answers:
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Answer:
(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz
(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz
Explanation:
(a) The fundamental, f₁, frequency is given as follows;
Where;
T = The tension in the string
μ = The linear density of the string
L = The length of the string
f₁ = The fundamental frequency = 560 Hz
If the tension in the string is increased by 15%, we will have;
Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz
(b) When the string length is decreased by one-third, we have;
The new length of the string, = 2/3·L
The value of the fundamental frequency will then be given as follows;
When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.
Answer:
A.The spring constant for B is one quarter of the spring constant for A.
Explanation:
If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.
As , and the 2 springs have the same mass
So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.