Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.
Explanation: To find the answer, we need to know more about the simple harmonic motion.
<h3>
What is simple harmonic motion?</h3>
- A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
- We have the equation of time period of a SHM as,
- Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
- We have to find the value of m,
Thus, we can conclude that, the mass m will be 0.625kg.
Learn more about simple harmonic motion here:
brainly.com/question/28045110
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Can someone help me with a question? At the local ski park, James earns 519.36 for working 32 hours. Create a double number line to identify the amount of money he would make for 1, 8 and 40 hours of work?
The weight changes but the mass will stay the same.
Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
