Physics, 24.01.2020 21:31 skypeltier

Aspring of constant k hangs from a ceiling with a mass m attached at the other end. the natural length of the hanging spring without the mass is l. a) draw this scenario, including forces and known or calculable displacements. b) what is the equilibrium length of the spring with the mass attached? c) suppose the spring is pulled down to create a displacement y from this equilibrium position and then released. write an expression y (t) describing its position as a function of time t. let "up" be the positive vertical direction. d) what is the period of oscillation for this simple harmonic oscillator? e) suppose you cut the spring in half. what would be the spring constant of each half-spring? (the two are identical, so they will have the same spring constant.) to determine this, redraw your diagram as two springs of equal length connected "in series," i. e., top to bottom. then indicate the forces acting on each of the three interesting bodies: the mass m and each halfspring of constant k’. apply hooke’s law to each half-spring individually and then to the two springs together. f) take the two half-springs from part e) and connect them "in parallel" (that is, side by side, hanging from the ceiling and each connected to the mass m). draw the new diagram and repeat parts a) - d)

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Answer from: chris3633

a) See part a of the attached picture.

There are two forces involved here:

- The weight of the block attached to the spring, of magnitude

W = mg

where m is the mass of the block and g = 9.8 m/s^2 is the acceleration due to gravity, pointing downward

- The restoring force of the spring,

F = kx

with k being the spring constant and x the displacement of the spring with respect to the equilibrium position, pointing upward

b) $L' = L+\frac{mg}{k}$

The spring is in equilibrium when the restoring force of the spring is equal to the weigth of the block:

kx = mg

where

x is the displacement of the spring with respect to the natural length of the spring, L. Solving for x,

$x=\frac{mg}{k}$

And since the natural length is L, the equilibrium length of the spring is

$L' = L+x=L+\frac{mg}{k}$

c) $y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)$

Let's assume that

y = 0

corresponds to the equilibrium position of the spring (which is stretched by an amount $L+\frac{mg}{k}$ ). If doing so, the vertical position of the mass at time t is given by

$y(t) = y cos(\omega t + \pi)$

where

$\omega = \sqrt{\frac{k}{m}}$ is the angular frequency

t is the time

y is the initial displacement with respect to the equilibrium position

$\pi$ is the phase shift, so that the position at time t=0 is negative:

y(0) = -y

So rewriting the angular frequency:

$y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)$

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

The period of oscillation is given by

$T=\frac{2\pi}{\omega}$

where

$\omega = \sqrt{\frac{k}{m}}$ is the angular frequency

Substituting $\omega$ , we find an expression for the period

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

e) k' = 2k (see part e of attached picture)

Here the two half springs of spring constant k' are connected in series, so the sum of the stretchings of the two springs is equal to the total stretching of the spring, x:

x=x_1 + x_2

Also, since the two springs are identical, their stretching will be the same:

x_1 = x_2 = x'

so we have

x=x'+x'=2x'

$x'=\frac{x}{2}$

Substituting x find in part (b),

$x'=\frac{mg}{2k}$ (1)

Hooke's law for each spring can be written as

F=k' x' (2)

where

F = mg (3)

is still the weight of the block

Using (1), (2) and (3) together, we find an expression for the spring constant k' of each spring

$mg=k'\frac{mg}{2k}\\k' =2k$

So, the spring constant of each half-spring is twice the spring constant of the original spring.

fa) See part f) of attached picture

This time we have the block hanging from both the two half-springs, each with spring constant k'. So at equilibrium, the weight of the block is equal to the sum of the restoring forces of the two springs:

k' x_1 + k' x_2 = mg

fb) $x'=\frac{L}{2}+\frac{mg}{2k'}$

For the two springs in parallel, the sum of the restoring forces of the two springs must be equal to the weight of the block:

$F_1 + F_2 = mg\\k_1 x_1 + k_2 x_2 = mg$

The two springs are identical, so they have same spring constant:

k_1 = k_2 = k'

So (1) can be rewritten as

k' (x_1 + x_2) = mg

And since the two springs are identical, their stretching x' is the same:

x_1 = x_2 = x'

so we can rewrite this as

k' (2x') = mg

$x'=\frac{mg}{2k'}$

and so the equilibrium length of each spring will be

$x'=\frac{L}{2}+\frac{mg}{2k'}$

fc) $y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)$

The system of two springs in parallel can be treated as a system of a single spring with equivalent spring constant given by

$k_{eq}=k_1 + k_2 = 2k'$

where k' is the spring constant of each spring.

So, let's assume again that

y = 0

corresponds to the equilibrium position as calculated in the previous part. If doing so, the vertical position of the mass at time t is given by:

$y(t) = y cos(\omega t + \pi)$

where this time we have

$\omega = \sqrt{\frac{k_{eq}}{m}}$ is the angular frequency of the system

t is the time

y is the initial displacement with respect to the equilibrium position

$\pi$ is the phase shift, which we put so that the position at time t=0 is negative:

y(0) = -y

If we rewrite the angular frequency,

$\omega = \sqrt{\frac{2k'}{m}}$

the position of the mass is

$y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)$

fd) $T=2 \pi \sqrt{\frac{m}{2k'}}$

Similarly to part d), the period of oscillation is

$T=\frac{2\pi}{\omega}$

where

$\omega = \sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{2k'}{m}}$ is the angular frequency

Substituting $\omega$ , we find

$T=2 \pi \sqrt{\frac{m}{2k'}}$

Aspring of constant k hangs from a ceiling with a mass m attached at the other end. the natural leng

Answer from: Quest

Ithink d is the right answer. hope this !

Answer from: Quest

show the wavelengths of a wave. show the crests of a wave.

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