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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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5 0
3 years ago
What is the De Broglie wavelength of an electron under 150 V acceleration?
yanalaym [24]

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron m=9.1\times 10^{-31}

Let the velocity of electron = v

Charge on the electron =1.6\times 10^{-19}C

plank's constant h =6.67\times 10^{-34}

According to energy conservation eV =\frac{1}{2}mv^2

v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec

Now we know that De Broglie wavelength \lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm

4 0
3 years ago
Read 2 more answers
Let S = { p q |p, q are prime numbers greater than 0} and E = {0, −2, 2, −4, 4, −6, 6, · · · } be the set of even integers. . Pr
Finger [1]

Answer:

prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S

Explanation:

Given that S = { p q |p, q are prime numbers greater than 0}

                    E = {0, −2, 2, −4, 4, −6, 6, · · · }

To prove  by constructing a bijection from S to E

detailed  solution attached below

After the bijection :

<em>prove that | S | = | E |</em> :  every element of S there is an Image on E , while not every element on E has an image on S

∴ we can say sets E and S are infinite sets

7 0
3 years ago
BTSAUDY5 NAME STARTS FROM THIS
otez555 [7]

Answer:

What's your question?

Explanation:

I'm not able to understand it....

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7 0
3 years ago
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Number pattern Write a recursive method called print Pattern() to output the following number pattern. Given a positive integer
lilavasa [31]

Answer:

See explaination

Explanation:

Code;

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void printNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

printNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

printNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

printNumPattern(num1, num2);

}

}

See attachment for sample output

3 0
3 years ago
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