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3 years ago

7

Write down the DE of a simple harmonic oscillator (for example, a mass-spring system) Draw a diagram to show a mass spring syste

m .

1 answer:

Sav [38]3 years ago

7 0

Answer:

let m be the mass of the object, K be the force constant and Fs be the force by the spring on the mass.

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1. Astronomical observatories have been available since ancient times, and many cultures set aside special sites for astronomica

nydimaria [60]

Answer:

e telescopes

Explanation:

may i be marked brainliest?

4 0

3 years ago

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

ankoles [38]

<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3 s

Substituting

s = ut + 0.5 at²

s = 0 x 3 + 0.5 x 9.81 x 3²

s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0

4 years ago

A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas

siniylev [52]

Answer:

3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

Frequency of oscillation (f) = 0.429 Hz

Let the mass be 'm' kg.

Now, we know that, a spring-mass system undergoes Simple Harmonic Motion (SHM). The frequency of oscillation of SHM is given as:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Rewrite the above equation in terms of 'm'. This gives,

$2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}$

Now, plug in the given values and solve for 'm'. This gives,

$m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg$

Therefore, the mass of the truck is 3331.5 kg.

3 0

3 years ago

Using a mass of 1.8 g and the volume displaced by the sample, calculate the sample's density. A) 0.17 g/mL B) 0.35 g/mL C) 0.60

fomenos

The answer to that probably would be C excuse me if I am wrong.

8 0

4 years ago

Read 2 more answers

A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).

yawa3891 [41]

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

$f = n\frac{v}{2L}$

Where

v = speed of sound

L = Length of the pipe

n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

Re-arrange to find L,

$f = n\frac{v}{2L}\\L = \frac{nv}{2f}$

The radius between the two frequencies would be 4 to 5,

$\frac{528Hz}{660Hz}= \frac{4}{5}$

$4:5$

Therefore the frequencies are in the ratio of natural numbers. That is

$4f = 528\\f = \frac{528}{4}\\f = 132Hz$

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

$L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m$

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

$L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m$

Therefore the length of the pipe in the second harmonic is 2.6m

7 0

3 years ago