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sammy [17]
3 years ago
11

Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e

nd of the beam is free and one end is fixed. The cross section has a base of 100 mm and height of 300 mm, length of the beam is 5.9 m, E=20.5x10^10 N/m2 and density of 7830 kg/m3. Write your answer in rad/sec with 2 decimal points.

Engineering
2 answers:
marshall27 [118]3 years ago
8 0

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

\Delta x=\frac{PL^3}{3EI}

Hence we can write

P=\frac{3EI\cdot \Delta x}{L^3}

Comparing with the standard spring equation F=kx we find the cantilever analogous to spring with k=\frac{3EI}{L^3}

Now the angular frequency of a spring is given by

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

where

'm' is the mass of the load

Thus applying values we get

\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}

\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec

Harman [31]3 years ago
8 0

Answer:

its a lil confusing

Explanation:

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Does anybody know what plane this is? i saw it the other day doing a low pass through my community
Arlecino [84]

Answer:

Airbus A340-313

Explanation:

it is what it is

3 0
2 years ago
For this problem, you may not look at any other code or pseudo-code (even if it is on the internet), other than what is on our w
sergiy2304 [10]

Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

while A[parent] < A[current] && parent != -1 do,

if A[parent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

current = parent

(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))

(b)

(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.

If above condition is not true then we will check for parent node and if it is less than current node then swap these.

pseudo code :-

current = i

// assuming parent of root is -1

parent is parent node of current node

while A[parent] < A[current] && parent != -1 do,

if A[grandparent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

swap(A[grandparent],A[parent])

current = grandparent

else if A[parent] < A[current]

swap(A[parent],A[current])

current = parent

(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)

so order would be log(n)/2

(iii) C++ code :-

#include<bits/stdc++.h>

using namespace std;

// function to return index of parent node

int parent(int i)

{

if(i == 0)

return -1;

return (i-1)/2;

}

// function to return index of grandparent node

int grandparent(int i)

{

int p = parent(i);

if(p == -1)

return -1;

else

return parent(p);

}

void shift_up(int A[], int n, int ind)

{

int curr = ind-1; // because array is 0-indexed

while(parent(curr) != -1 && A[parent(curr)] < A[curr])

{

int g = grandparent(curr);

int p = parent(curr);

if(g != -1 && A[g] < A[curr])

{

swap(A[curr],A[p]);

swap(A[p],A[g]);

curr = g;

}

else if(A[p] < A[curr])

{

swap(A[p],A[curr]);

curr = p;

}

}

}

int main()

{

int n;

cout<<"enter the number of elements :-\n";

cin>>n;

int A[n];

cout<<"enter the elements of array :-\n";

for(int i=0;i<n;i++)

cin>>A[i];

int ind;

cout<<"enter the index value :- \n";

cin>>ind;

shift_up(A,n,ind);

cout<<"array after shift up :-\n";

for(int i=0;i<n;i++)

cout<<A[i]<<" ";

cout<<endl;

}

Explanation:

8 0
3 years ago
A technique for resolving complex repetitive waveforms into sine or cosine waves and a DC component is known as:
tatiyna

Answer:

(A) Fourier Analysis

Explanation:

Fourier Analysis  

It is the form of study of the way a general functions can be represented via the sum of the simple trigonometric functions .

It is named after Joseph Fourier , who represented a function as a sum of its trigonometric functions and it simplifies the study of the heat transfer .

Hence ,  

The technique for resolving the complex repetitive waveforms into the sine or the cosine waves and the DC component is known as the Fourier Analysis .

7 0
4 years ago
What is the difference between a stepped and a non-stepped ECT circuit?<br> ​
galina1969 [7]

Answer:

A stepped circuit is designed in the engine coolant temperature (ECT) sensor within a powertrain control module (PCM) to increase the sensor's accuracy. A simple non-stepped ECT sensor circuit has a specific resistance which increases or decreases according to the changes in engine coolant temperature.

Explanation:

5 0
3 years ago
Sometimes, we need an amplifier with an accurate gain. Even using 1% tolerance resistors may not be sufficient to provide the re
vlabodo [156]

Answer:

The value of R_low is 13.1 kΩ while that of R_high is 36.8 kΩ

Explanation:

The diagram is as given as

The circuit is an inverting op amp for which the gain is given as

A_v=\dfrac{R_2+R}{R_1+R_z}

Here R2 is given as 450 kΩ.

R1 is given as 47 kΩ.

The gain is given as

A_v=-10\pm 0.5*\dfrac{10}{100}\\A_v=-10\pm 0.05

For the Lowest value of R the value of the internal resistance rz is 0 and the gain is -10+0.05 so

A_v=-\dfrac{R_2+R}{R_1+R_z}\\-10+0.05=-\dfrac{450+R_l_o_w}{47+0}\\-9.95=-\dfrac{450+R_l_o_w}{47+0}\\R_l_o_w=13.1

So the value of R_low is 13.1 kΩ

For the Highest value of R the value of the internal resistance rz is 1 kΩ and the gain is -10-0.05 so

A_v=-\dfrac{R_2+R}{R_1+R_z}\\-10-0.05=-\dfrac{450+R_{high}}{47+1}\\-10.05=-\dfrac{450+R_{high}}{47+1}\\R_{high}=36.8

So the value of R_high is 36.8 kΩ

5 0
3 years ago
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