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.