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3 years ago

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A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple

harmonic motion with a period of 1s. If that same mass–spring system is displaced from equilibrium by 10cm instead, what will its period be in this case?(A) 0.5s(B) 2s(C) 1s(D) 1.4s

1 answer:

aliina [53]3 years ago

5 0

Answer:

T= 1 s

Explanation:

Given that

When x= cm ,T= 1

we know that time period of spring mas system given as

$T=2\pi \sqrt{\dfrac{m}{k}}$

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

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To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

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Read 2 more answers

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we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

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