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2 years ago

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3. A ball of mass 2.0 kg is attached to a vertical spring. The spring constant is 196 N/m. What is the period if it were to be i

n SHM?
-0.063
-0.63 5
-6.3 s
-63 s
PLEASEHELPME

1 answer:

Reil [10]2 years ago

4 0

The period of the ball in simple harmonic motion is 0.634 s.

<h3>What is period?</h3>

Period can be defined as the time ( in seconds) taken for a body to complete one oscillation.

To calculate the period of the ball in SHM, we use the formula below.

Formula:

T = 2π√(m/k)............... Equation 1

Where:

T = Period of the ball
m = mass of the ball
k = spring constant
π = pie

From the question,

Given:

m = 2 kg
k = 196 N/m
π = 3.14

Substitute these values into equation 1

T = 2(3.14)√(2/196)
T = 6.28√(0.010204)
T = 0.634 s

Hence, The period of the ball in simple harmonic motion is 0.634 s.

Learn more about period here: brainly.com/question/21924087

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$-\frac{b}{2m}t=ln\frac{A_2}{A_1}$

$lnx=y\implies x=e^y$

Substitute the values

$-\frac{b}{2\times 0.046}\times 4.55=ln\frac{0.12}{0.296}$

$\frac{2\times 0.046}{4.55b}=ln\frac{0.296}{0.12}$

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