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4 years ago

13

If the mass attached to a spring is increased and the spring constant is decreased, does the proof of the oscillating does the m

otions increase, decrease or stay the same

1 answer:

Fofino [41]4 years ago

7 0

Answer:

Will increase

Explanation:

The period of an oscillating motion is the time it takes for the system to make one complete oscillation.

The period of a spring-mass system is given by the equation:

$T=2\pi \sqrt{\frac{m}{k}}$

where:

k is the spring constant

m is the mass attached to the spring

In this problem:

- The mass of the system is increased

- The spring constant is decreased

We observe that:

- The period of the system is proportional to the square root of the mass: so as the mass increases, the period will increase as well

- The period is inversely proportional to the square root of the spring constant: so as the constant decreases, the period will increase

Therefore, this means that in this case, the period will increase.

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A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou

zloy xaker [14]

Answer:

1.5288 J

Explanation:

We can use the Energy conversion law

Energy can not be made or destroyed. It can only be converted to another form of energy

Potential Energy (PE) is the energy an object has due to its height of the location

<em>PE = mass * gravitational acceleration * height from the ground</em>

Simply here , lemming had a PE and KE also.

(g = 9.8 m/s²)

PE = 0.0780 * 9.8 *2

= 1.5288 J

4 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

4 years ago

Contrast situations where work is done with different amounts of force to situations where no work is done such as standing stil

uranmaximum [27]

Picking up a sheet of paper . . . work done with small force
Picking up a glass of water . . . work done with moderate force
Picking up a huge boulder . . . work done with a great tremendous force
=================================
Standing still . . .
Holding your tongue out as far as it will go . . .
Holding your arm over your head for 3 days . . .
Holding a huge boulder motionless over your head . . .
Pushing on a brick wall . . .
Pushing as hard as you can against a truck with the wheels locked . . .
. . . . . No work done at all, because the force doesn't move through a distance.

<u>Work done = (force) times (distance)</u>

If the force doesn't move, then the distance is zero, and the work done is zero.

5 0

3 years ago

What element from the periodic table rhymes with extreme

jonny [76]

Halite or sulfur or gold or silver

8 0

4 years ago

An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k

skelet666 [1.2K]

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ( $U_{e}$ ), in joules, is equal to maximum translational kinetic energy ( $K$ ), in joules:

$U_{e} = K$

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$ (1)

Where:

$k$ - Spring constant, in newtons per meter.

$A$ - Amplitude, in meters.

$m$ - Object mass, in kilograms.

$v$ - Speed of the object at equilibrium, in meters per second.

If we know that $k = 450\,\frac{N}{m}$ , $m = 0.25\,kg$ and $v = 0.3\,\frac{m}{s}$ , then the amplitude of the motion is:

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$k\cdot A^{2} = m\cdot v^{2}$

$A = v\cdot \sqrt{\frac{m}{k} }$

$A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }$

$A \approx 0.274\,m$

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object ( $E$ ), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ( $k = 450\,\frac{N}{m}$ , $A \approx 0.274\,m$ )

$E = U_{e}$

$E = \frac{1}{2}\cdot k\cdot A^{2}$

$E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}$

$E = 16.892\,J$

The total energy of the object at any point of its motion is 16.892 joules.

7 0

3 years ago