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3 years ago

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A 2.40 kg ball is attached to an unknown spring and allowed to oscillate. The figure below shows a graph of the ball’s position

�� as a function of time . What are the oscillation’s: a. period; b. frequency; c. angular frequency; d. amplitude; and e. What is the force constant of the spring?

1 answer:

irga5000 [103]3 years ago

4 0

(a) The period of the oscillation is 0.8 s.

(b) The frequency of the oscillation is 1.25 Hz.

(c) The angular frequency of the oscillation is 7.885 rad/s.

(d) The amplitude of the oscillation is 3 cm.

(e) The force constant of the spring is 148.1 N/m.

The given parameters:

<em>Mass of the ball, m = 2.4 kg</em>

<em />

From the given graph, we can determine the missing parameters.

The amplitude of the wave is the maximum displacement, A = 3 cm

The period of the oscillation is the time taken to make one complete cycle.

T = 0.8 s

The frequency of the oscillation is calculated as follows;

$f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz$

The angular frequency of the oscillation is calculated as follows;

$\omega = 2\pi f\\\\\omega = 2\pi \times 1.25\\\\\omega = 7.855 \ rad/s$

The force constant of the spring is calculated as follows;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\ k = \omega ^2 m\\\\k = (7.855)^2 \times 2.4\\\\k = 148.1 \ N/m$

Learn more about general wave equation here: brainly.com/question/25699025

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3 years ago

17. A 1350 g projectile is launched with a force of 150 N. The length of the firing arm is 1.25 m.

creativ13 [48]

a) The exit velocity of the projectile is 16.7 m/s

b) The maximum height achieved by the projectile is 14.2 m

c) The total time of flight is 3.40 s

d) The distance covered by the projectile is 46.5 m

Explanation:

a)

We solve this first part of the problem by applying the work-energy theorem, which states that the work done on the projectile is equal to the gain in kinetic energy of the projectile. Mathematically:

$W=Fd = \frac{1}{2}mv^2-\frac{1}{2}mu^2=\Delta K$

where:

F = 150 N is the force applied

d = 1.25 m is the displacement of the projectile (the length of the firing arm)

m = 1350 g = 1.35 kg is the mass of the projectile

u = 0 is the initial velocity of the projectile

v is the exit velocity of the projectile

Solving for v, we find:

$v=\sqrt{\frac{2Fd}{m}}=\sqrt{\frac{2(150)(1.25)}{1.35}}=16.7 m/s$

b)

Assuming the projectile is fired vertically upward, then the initial kinetic energy of the projectile as soon as he leaves the cannon is fully converted into gravitational potential energy as it reaches the top of its trajectory. So we can write:

$K_i = U_f$

$\frac{1}{2}mv^2=mgh$

where:

$K_i$ is the initial kinetic energy

$U_f$ is the final potential energy

m = 1350 g = 1.35 kg is the mass of the projectile

v = 16.7 m/s is the velocity at which the projectile leaves the cannon

$g=9.8 m/s^2$ is the acceleration of gravity

h is the maximum height reached by the projectile

And solving for h, we find

$h=\frac{v^2}{2g}=\frac{(16.7)^2}{2(9.8)}=14.2 m$

c)

Assuming the projectile is launched vertically upward, then the total time of flight is twice the time it takes for reaching the maximum height. This time can be found by using the following suvat equation:

$v=u-gt$

where:

u = 16.7 m/s is the initial velocity

$g=9.8 m/s^2$ is the acceleration of gravity

t is the time

The projectile reaches the maximum height when the vertical velocity becomes zero, so when v = 0. Therefore, substituting,

$0=u-gt\\t=\frac{u}{g}=\frac{16.7}{9.8}=1.70 s$

So, the total time of flight is

$T=2t=2(1.70)=3.40 s$

d)

The motion of a projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

First we have to analyze the vertical motion, to find the time of flight of the apple. We can do it by using the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where

s = -27 m is the vertical displacement of the apple

$u_y=u sin \theta = (16.7)(sin 42^{\circ})=11.2 m/s$ is the initial vertical velocity

t is the time if flight

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting we have:

$-27=11.2t - 4.9t^2\\4.9t^2-11.2t+27=0$

which has two solutions:

t = -1.46 s (negative, we discarde)

t = 3.75 s (this is our solution)

Now we can analyze the horizontal motion: the projectile moves horizontally with a constant velocity of

$v_x = u cos \theta = (16.7)(cos 42^{\circ})=12.4 m/s$

So, the distance it covers during its fall is given by

$d=v_x t=(12.4)(3.75)=46.5 m$

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