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n200080 [17]
4 years ago
15

A 0.90- kg mass attached to a vertical spring of force constant 170 N/m oscillates with a maximum speed of 0.20 m/s . Find the f

ollowing quantities related to the motion of the mass.
Find the period.

Find the amplitude.

Find the maximum magnitude of the acceleration.
Express your answer using two significant figures.
Physics
1 answer:
Alex4 years ago
3 0

Explanation:

Given that,

The mass of the object, m = 0.9 kg

Force constant, k = 170 N/m

Maximum speed of the object, v = 0.2 m/s

Solution,

(a) The angular frequency of the object is given by :

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

\omega=\sqrt{\dfrac{170}{0.9}}

\omega=13.74\ rad/s

The time period is given by :

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

T=\dfrac{2\pi}{13.74}

T = 0.45 seconds

(b) The maximum velocity of the object in shm is given by :

v=A\omega

Amplitude,

A=\dfrac{v}{\omega}

A=\dfrac{0.2}{13.74}

A = 0.0145 m

(c) The maximum acceleration of the object is given by :

a=\omega^2A

a=(13.74)^2\times 0.0145

a=2.73\ m/s^2

Therefore, this is the required solution.

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Alain Aspect, David Baltimore, Allen Bard, and Timothy Berners - Lee

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4 years ago
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A tall cylinder contains 25 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t
Fed [463]

Answer: 377.3\ kPa

Explanation:

Given

Water column height h=25\ cm

After oil is poured, the total height becomes h'=40\ cm

Pressure at the bottom will be the sum due to the water and oil column

Suppose the density of the oil is \rho=900\ kg/m^3

Pressure at the bottom

\Rightarrow P=10^3\times g\times 25+900\times g\times 15\\\Rightarrow P=100g[250+135]\\\Rightarrow P=3773\times 100\ Pa\\\Rightarrow P=377.3\ kPa

4 0
3 years ago
A damping force affects the vibration of a spring so that the displacement of the spring is given by y = e−4t(cos 2t + 3 sin 2t)
astraxan [27]

Answer:

Average velocity = dy/dt = 4sin2t(2t+3)-4cos2t(6t+1)e^-4t(cos 2t + 3 sin 2t).

Explanation:

Velocity is defined as the rate of change in displacement.

Velocity = change in displacement/time taken

Given the displacement of the string as;

y = e^−4t(cos 2t + 3 sin 2t).

To get the average velocity, we will find the derivative of the displacement with respect to time.

Using function of a function to solve this;

Let u = 4t(cos 2t + 3 sin 2t)... (1)

y = e^-u... (2)

Differentiating both functions with respect to their variables we have;

dy/du = -e^-u

du/dt is gotten using the product rule to have;

du/dt = 4t(-2sin2t+6cos2t)+4(cos2t+3sin2t)

Opening up the bracket we have;

du/dt = -8tsin2t+24tcos2t+4cos2t+12sin2t

Collecting like terms;

-8tsin2t+12sin2t+24tcos2t+4cos2t

du/dt = -4sin2t(2t-3)+4cos2t(6t+1)

dy/dt = dy/du × du/dt

dy/dt = -e^-u × -4sin2t(2t+3)+4cos2t(6t+1)

Substituting u = 4t(cos 2t + 3 sin 2t) into dy/dt, we will have;

dy/dt = -e^-4t(cos 2t + 3 sin 2t) × -4sin2t(2t+3)+4cos2t(6t+1)

dy/dt = 4sin2t(2t+3)-4cos2t(6t+1)e^-4t(cos 2t + 3 sin 2t).

The average velocity of the wave function therefore give us;

dy/dt = 4sin2t(2t+3)-4cos2t(6t+1)e^-4t(cos 2t + 3 sin 2t).

4 0
3 years ago
In the attached position versus time graph what is the magnitude of average velocity for the entire 19 seconds of the motion. An
evablogger [386]

According to the plot, the positions at time <em>t</em> = 0 s and <em>t</em> = 19 s are -1 m and -2 m, respectively. So the average velocity for the 19-s interval is

v_{\rm ave} = \dfrac{-2\,\mathrm m - (-1\,\mathrm m)}{19\,\mathrm s} = -\dfrac1{19}\dfrac{\rm m}{\rm s}\approx \boxed{-0.05\dfrac{\rm m}{\rm s}}

8 0
3 years ago
How to add an answer
netineya [11]
You cannot add an answer if two people are already answering!
7 0
3 years ago
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