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:
1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x 
J
2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x J
Explanation:
given information:
= 3 nC = 3 x 
C
= 2 nC = 2 x C
r = 50 cm = 0.5 m
the electric potential energy of the electron when it is at the midpoint
potential energy of the charge, F
F = k 
where
k = constant (8.99 x 
)
electron charge, 
= - 1.6 x 
C
since it is measured at the midpoint,
r = 
= 0.25 m
thus,
F = 
= k
+ k
= 
(
)
= (8.99 x 
)( - 1.6 x )(3 x +2 x )/0.25
= - 2.9 x J
the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge
= 10 cm = 0.1 m
= 0.5 - 0.1 = 0.4 m
F = k + k
= 
(
+
)
= (8.99 x )( - 1.6 x )(3 x /0.1+2 x /0.4)
= - 5.04 x J
the answer is (A) the movement of the magnet relative to the coil
Form concentric circles around the wire
