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SCORPION-xisa [38]
3 years ago
11

For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time

iv) Frequency v) On a diagram show the positions of max and min values for Acceleration and Velocity and show using the equations why this is the case.

Engineering
1 answer:
max2010maxim [7]3 years ago
4 0

Answer with Explanation:

The general equation of simple harmonic motion is

x(t)=Asin(\omega t+\phi)

where,

A is the amplitude of motion

\omega is the angular frequency of the motion

\phi is known as initial phase

part 1)

Now by definition of velocity we have

v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )

part 2)

Now by definition of acceleration we have

a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )

part 3)

The angular frequency is related to Time period 'T' asT =\frac{2\pi }{\omega }

where

\omega is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is A\omega ^{2} and A\omega respectively.

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Answer:

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v  - u  =  a * t

v - a * t = u

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3. How can statistical analysis of a dataset inform a design process?
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A vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i
Readme [11.4K]

Answer:11.602 KW

Explanation:

mass of vehicle\left ( m\right )=1200 kg

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Net force to overcome by engine is

F=Resistance + sin component of weight

F=600+mgsin\theta

Where tan\theta =[tex]\frac{1}{50}

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F=600+1200\times 9.81\times sin\left ( 1.1457\right )

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4 0
4 years ago
In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery
Dima020 [189]

Answer:

  248.756 mV

  49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

  voltage: (100 V)(1000/(1000 +1000) = 50 V

  impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

  voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

  impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

__

The open-circuit voltage is the difference between these terminal voltages:

  (50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

__

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

  (50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0
3 years ago
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