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a block oscillating on a spring has an amplitude of 20cm. what will be the amplitude if the maximum kinetic energy is doubled? 7. a block oscillating on a spring has a maximum kinetic energy of 2.01. what will be the maximum kinetic energy if the ampli tude is doubled? explain.

User Terry Low
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Answer:

To answer this question, we can use the equation for the maximum kinetic energy of a block oscillating on a spring:

K_max = (1/2) * m * ω^2 * A^2

where m is the mass of the block, ω is the angular frequency of the oscillation, and A is the amplitude of the oscillation.

If we double the maximum kinetic energy while keeping all other variables constant, we get:

2 * K_max = (1/2) * m * ω^2 * A_new^2

Dividing both sides by the original equation for K_max, we get:

2 = (A_new / A)^2

Taking the square root of both sides, we get:

√2 = A_new / A

Therefore, the new amplitude (A_new) is equal to the original amplitude (A) multiplied by the square root of 2. Plugging in the given value of 20 cm for the original amplitude, we get:

A_new = √2 * 20 cm ≈ 28.3 cm

So the new amplitude would be approximately 28.3 cm.

To answer this question, we can use the same equation for the maximum kinetic energy of a block oscillating on a spring:

K_max = (1/2) * m * ω^2 * A^2

If we double the amplitude while keeping all other variables constant, we get:

K_max_new = (1/2) * m * ω^2 * (2A)^2

Simplifying this expression, we get:

K_max_new = 2 * K_max

Therefore, the new maximum kinetic energy (K_max_new) is equal to the original maximum kinetic energy (K_max) multiplied by 2. Plugging in the given value of 2.01 for the original maximum kinetic energy, we get:

K_max_new = 2 * 2.01 = 4.02

So the new maximum kinetic energy would be 4.02.

User Xmetal
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