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Alex777 [14]
3 years ago
11

A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates wi

th a maximum amplitude of 10 mm. What is its natural frequency?
Engineering
1 answer:
____ [38]3 years ago
6 0

Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

V_{max}=\omega A

now by putting the values

V_{max}=\omega A

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

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Select four examples of fluid or pneumatic power systems.
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Read 2 more answers
A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
When a variable is stored in memory, it is associated with an address. To obtain the address of a variable, the & operator c
liubo4ka [24]

Answer:

Explanation:

1) C program file addressOfScalar.c

#include <stdio.h>

int main()

{

//intialize a char variable, print its address and the next address

char charvar = 'a';

printf("address of charvar = %p\n", (void *)(&charvar));

printf("address of charvar - 1 = %p\n", (void *)(&charvar - 1));

printf("address of charvar + 1 = %p\n", (void *)(&charvar + 1));

//intialize a int variable, print its address and the next address

int intvar = 1;

printf("address of intvar = %p\n", (void *)(&intvar));

printf("address of intvar - 1 = %p\n", (void *)(&intvar - 1));

printf("address of intvar + 1 = %p\n", (void *)(&intvar + 1));

}

In C programming language, an int variable takes 4 bytes of memory. So any arithmetic on integer address, always considers it as 4 bytes of data. So intvar-1 refers to a location 4 bytes before intvar's address and intvar+1 refers to 4 bytes after intvar's address.

3 0
3 years ago
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