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3 years ago

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A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , wher

e x is in cm and t is in seconds. What is the frequency of the oscillation?

2 answers:

zhuklara [117]3 years ago

7 0

Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

$x(t) = Acos(\omega t - \phi)$

where A is the amplitude, $\omega = 2\pi f$ is the angular frequency and $\phi$ is the initial phase.

Since our body has an equation of x = 5cos(π t + π/3) we can equate $\omega = \pi$ and solve for frequency f

$2\pi f = \pi$

f = 1/2 Hz

IRISSAK [1]3 years ago

6 0

<h2>Answer:</h2>

0.5Hz

<h2>Explanation:</h2>

The general equation of the displacement, x, of a body undergoing simple harmonic motion at a given point in time (t) is given by;

x = A cos (ωt ± ∅) --------------------------(i)

where;

A = amplitude of the wave

ω = angular velocity of the wave

∅ = phase constant of the wave

<em>From the question;</em>

x = 5cos(π t + π/3) -----------------------------(ii)

<em>Comparing equations (i) and (ii), the following deductions among others can be made;</em>

A = 5cm

ω = π

<em>But the angular velocity (ω) of the wave is related to its frequency (f) as follows;</em>

ω = 2 π f --------------------(iii)

<em>Substitute the value of ω = π into equation (iii) as follows;</em>

π = 2 π f

<em>Divide through by π;</em>

1 = 2f

<em>Solve for f;</em>

f = 1/2

f = 0.5

Frequency (f) is measured in Hz. Therefore, the frequency of the oscillation is 0.5Hz

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3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

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Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

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$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

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c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

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3 years ago

The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.55 g of conden

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Answer:

m = 62.14 g

Explanation:

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