Question

You stretch a spring ball system 0.552 m away from its equilibrium point and watch it oscillate. You find that the system's angular frequency is 1.42 rad/sec. What is the maximum speed of the ball?

Answer 1

Answer:

The maximum speed of ball is 0.784 m/s.

Explanation:

Given that,

Amplitude = 0.552 m

Angular frequency = 1.42 rad/s

We need to calculate the maximum speed of ball

Using formula of speed

$v=A\omega$

Where, A = amplitude

$\omega$ = angular frequency

Put the value into the formula

$v=0.552\times1.42$

$v=0.784\ m/s$

Hence, The maximum speed of ball is 0.784 m/s.

Answer 2

The maximum speed of the ball is 0.784 m/s.

To calculate the maximum speed of the ball, we use the formula below.

Formula:

• s = ωr............... Equation 1

Where:

• s = maximum speed of the ball
• ω = angular frequency
• r = distance of the ball from its equilibrium position

From the question,

Given:

• ω = 1.42 rad/s
• r = 0.552 m

Substitute these values into equation 1

• s = 0.552(1.42)
• s = 0.784 m/s.

Hence, The maximum speed of the ball is 0.784 m/s.

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