Questions: A mass suspended by a spring that obeys Hooke's law (F=-k) will stretch the spring a distance x. In equilibrium, the spring pulls to balance the weight, mg, of the mass. The spring constant is "k", and the acceleration of gravity is "g". When the spring is stretched from equilibrium and released, the mass will oscillate up and down at a frequency f=1 / 2π√(k / m). Imagine a coiled spring that is extended a distance 0.09 by a mass hanging from it: it is then pulled down more and released so that it oscillates up and down. How long does it take to go up to its maximum and back down to its point of release? Give your answer in seconds but do not include the units when you enter it here on Blackboard. Hint: It does not depend on the mass!

A mass suspended by a spring that obeys Hooke's law (F=-k) will stretch the spring a distance x. In equilibrium, the spring pulls to balance the weight, mg, of the mass. The spring constant is "k", and the acceleration of gravity is "g". When the spring is stretched from equilibrium and released, the mass will oscillate up and down at a frequency f=1 / 2π√(k / m). Imagine a coiled spring that is extended a distance 0.09 by a mass hanging from it: it is then pulled down more and released so that it oscillates up and down. How long does it take to go up to its maximum and back down to its point of release? Give your answer in seconds but do not include the units when you enter it here on Blackboard. Hint: It does not depend on the mass!

Transcript text: A mass suspended by a spring that obeys Hooke's law ( $F=-\mathrm{k}$ ) will stretch the spring a distance $x$. In equilibrium, the spring pulls to balance the weight, mg , of the mass. The spring constant is " k ", and the acceleration of gravity is " g ". When the spring is stretched from equilibrium and released, the mass will oscillate up and down at a frequency $f=1 / 2 \pi \sqrt{ }(\mathrm{k} / \mathrm{m})$. Imagine a coiled spring that is extended a distance 0.09 by a mass hanging from it: it is then pulled down more and released so that it oscillates up and down. How long does it take to go up to its maximum and back down to its point of release? Give your answer in seconds but do not include the units when you enter it here on Blackboard. Hint: It does not depend on the mass!

Solution

Solution Steps

Step 1: Understanding the Problem

We need to determine the time it takes for a mass-spring system to oscillate from its maximum displacement back to the point of release. The system follows Hooke's law and oscillates with a frequency given by $ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $.

Step 2: Determine the Period of Oscillation

The period $ T $ of the oscillation is the reciprocal of the frequency: \[ T = \frac{1}{f} = 2\pi \sqrt{\frac{m}{k}} \]

Step 3: Calculate the Time for Half an Oscillation

The time it takes to go from the maximum displacement to the point of release is half of the period: \[ \text{Time} = \frac{T}{2} = \frac{2\pi \sqrt{\frac{m}{k}}}{2} = \pi \sqrt{\frac{m}{k}} \]

Step 4: Simplify the Expression

Since the problem states that the time does not depend on the mass, we can use the given hint to simplify our calculations. The time for half an oscillation is: \[ \text{Time} = \pi \sqrt{\frac{m}{k}} \]

Final Answer

\[ \boxed{\pi \sqrt{\frac{m}{k}}} \]

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