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Alex_Xolod [135]
3 years ago
10

When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the

period of oscillation, T , neglecting the mass of the spring itself.
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=\sqrt{\dfrac{k}{m}}

Where, m = mass

k = spring constant

Put the value into the formula

\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}

\omega=4.74\ rad/s

We need to calculate the period of oscillation,

Using formula of time period

T=\dfrac{2\pi}{\omega}

Put the value into the formula

T=\dfrac{2\pi}{4.74}

T=1.33\ sec

Hence, The period of oscillation is 1.33 sec.

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enyata [817]

If you fly at 100 miles per hour for a time of 2 hours and 30 minutes you will be a: 250 miles far

The formula and procedure we will use to solve this exercise is:

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Information about the problem:

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Applying the distance formula we have that:

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<h3>What is velocity?</h3>

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

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8 0
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The batter swings and misses the 40 m/s (90 mph) fastball, and the ball (mass 150 grams) ends up at rest in the catcher's mitt.
Nadusha1986 [10]

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5 0
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(I can't really be sure of anything, because whenever you show a formula
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