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Zina [86]
4 years ago
14

A 100-coil spring has a spring constant of 540 N/m. It is cut into four shorter springs, each of which has 25 coils. One end of

a 25-coil spring is attached to a wall. An object of mass 76 kg is attached to the other end of the spring, and the system is set into horizontal oscillation. What is the angular frequency of the motion? Number Entry field with incorrect answer now contains modified data Units Entry field with correct answer
Physics
1 answer:
amid [387]4 years ago
6 0

Answer:

w = 5.3311 rad/sec

Explanation:

n = 100 has a k = 540 N/m

k depends on the number of coils by:

k = \frac{G*d^4}{8*n*D^3}

By the design equation we see that the spring stiffness k has an inverse relationship with number of coils n.

Hence, when n = 25 coils ; k = 4* 540 = 2160 N/m

The relationship between angular frequency and k is:

w = \sqrt{\frac{k}{m} }\\\\Hence,\\\\w = \sqrt{\frac{2160}{76} }\\\\w = 5.3311 rad/sec

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The frequency and amplitude of the SHM beam is 0.8 Hz and 0.098 m. The frequency of the SHM wave when gravel falls is 0.8 Hz and the amplitude of subsequent SHM beam is 0.4m.

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Mass of the sack = 175 Kg

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Frequency of the beam = F = 0.60 cycles/s

The formula for frequency of oscillation =

= f = (1/2π) X √(k/m)

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= x = mg / k

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= X = 0.289 m

Now, since we know that the gravel falls, thus frequency = f =

= f = (1/2π) X √(k/m)

= f= (1/ 2 X 3.14) X √ 5685.37 / 225

= f = 0.8 Hz

(b) Assuming that the spring is stretched, x = mg/k =

= x = (225 X 9.8) / 5685.37

= x = 0.3878 m

Thus, the amplitude of the sack = A = 0.3878 - 0.289

= A = 0.098 m

(c) If the gravel falls, the speed is maximum hence speed = s =

= s = A X √(k/m)

= s = 0.4 X √(5685.37/400)

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The frequency = f' =

= f' = (1/2π) X √(k/m)

= f' = (1/2 X 2.14) X √(5685.37/225)

= f' = 0.8 Hz

(d) New amplitude = A' =

= A' = 0.38 + 0.038   (after calculating the new distance)

= A' = 0.4 m

To know more about Spring:

brainly.com/question/15850235

#SPJ4

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