❤this is what i found hope it explains
Answer:
a) A = 4.0 m
, b) w = 3.0 rad / s
, c) f = 0.477 Hz
, d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1
²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
Answer: 3 m/s
Explanation:
The average velocity is simply the change in displacement over the specified time interval. I’m this case, we have a total of four intervals,which represent 4 seconds. Dividing 12 m by 4s gives us 3 m/s
The source of the sound it moving closer to you because the wavelength is shortened. The shorter the wavelength, the higher the frequency, the higher the pitch
This is how far I got:
![1. \ \frac{1}{2} m_1 v_1^{2} + \frac{1}{2} m_2 v_2^{2} = m_1gh \\ \\2.\ m_1v_1 + m_2v_2 = 0 \\ \\ for \ v_1 = 4 \\ \\3.\ v_2 = -4 \frac{m_1}{m_2}](https://tex.z-dn.net/?f=%201.%20%5C%20%5Cfrac%7B1%7D%7B2%7D%20m_1%20v_1%5E%7B2%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20m_2%20v_2%5E%7B2%7D%20%3D%20m_1gh%20%5C%5C%20%5C%5C2.%5C%20m_1v_1%20%2B%20m_2v_2%20%3D%200%20%5C%5C%20%5C%5C%20for%20%5C%20v_1%20%3D%204%20%5C%5C%20%5C%5C3.%5C%20v_2%20%3D%20-4%20%5Cfrac%7Bm_1%7D%7Bm_2%7D%20)
put equation 3. into 1. together with h = 5:
4.
![\frac{m_1}{m_2} = \frac{5}{8} g -1](https://tex.z-dn.net/?f=%20%5Cfrac%7Bm_1%7D%7Bm_2%7D%20%3D%20%5Cfrac%7B5%7D%7B8%7D%20g%20-1)
when I double m₁ the ratio becomes:
5.
![\frac{2m_1}{m_2} = \frac{5}{4} g -2](https://tex.z-dn.net/?f=%20%5Cfrac%7B2m_1%7D%7Bm_2%7D%20%3D%20%5Cfrac%7B5%7D%7B4%7D%20g%20-2)
when I put these to ratios back into equation nr 1 and nr2, I find the ratio:
![\frac{new\ v_1}{old\ v_1} = 0.737](https://tex.z-dn.net/?f=%20%5Cfrac%7Bnew%5C%20v_1%7D%7Bold%5C%20v_1%7D%20%3D%200.737)
and new v₁ = 2.95
I'm not sure this is correct.