Physics, 18.11.2020 17:10 mchillery1028

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again?

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Answer from: faizanaj3978

The time taken for the spring to reach new equilibrium position is 0.26 s.

The given parameters;

mass, m = 1.65 kgextension of the string, x = 0.26 mdisplacement with time x(t) = 0.13

The general wave equation is given as;

$y(t) = A\ cos(\omega t \ + \phi)$

The angular frequency is given as follows;

$\omega = \sqrt{\frac{k}{m} } \\\\$

The spring constant is calculated as;

$F = mg\\\\kx = mg\\\\k = \frac{mg}{x} \\\\k = \frac{1.65 \times 9.8}{0.26} \\\\k = 62.2 \ N/m$

The angular frequency is calculated as follows;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{62.2}{1.65} }\\\\\omega = 6.14 \ rad/s$

The time taken for the spring to reach new equilibrium position is calculated as follows;

$y(t) = A \ cos(\omega t)\\\\0 = 0.13 \times cos(6.14t)\\\\6.14t = cos^{-1}(0)\\\\6.14t = 1.57 \ rad\\\\t = \frac{1.57 \ rad}{6.14 \ rad/s} \\\\t = 0.26 \ s$

Thus, the time taken for the spring to reach new equilibrium position is 0.26 s.

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Answer from: davisnaziyahovz5sk

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

y(t) - Position of the mass as a function of time, measured in meters.

- Amplitude, measured in meters.

- Spring constant, measured in newtons per meter.

- Mass of the block, measured in kilograms.

- Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

- Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for . Given that cosine is a periodic function, we are only interested in the least value of such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Answer from: Quest

a) cold, dry

explanation:

continental polar masses are cold and dry air masses because the cold air cannot hold much moisture. as they move into warmer air, they can trigger precipitation.

Answer from: Quest

0.11 m/s and 0.28 m/s should be correct

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A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m...