The object with the mass ok 1kg will move more quickly because it is lighter than the 100kg object
Answer:
x(t) = d*cos ( wt )
w = √(k/m)
Explanation:
Given:-
- The mass of block = m
- The spring constant = k
- The initial displacement = xi = d
Find:-
- The expression for displacement (x) as function of time (t).
Solution:-
- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.
- We apply the Newton's equation of motion in horizontal direction.
F = ma
-kx = ma
-kx = mx''
mx'' + kx = 0
- Solve the Auxiliary equation for the ODE above:
ms^2 + k = 0
s^2 + (k/m) = 0
s = +/- √(k/m) i = +/- w i
- The complementary solution for complex roots is:
x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]
- The given initial conditions are:
x(0) = d
d = [ A*cos ( 0 ) + B*sin ( 0 ) ]
d = A
x'(0) = 0
x'(t) = -Aw*sin (wt) + Bw*cos(wt)
0 = -Aw*sin (0) + Bw*cos(0)
B = 0
- The required displacement-time relationship for SHM:
x(t) = d*cos ( wt )
w = √(k/m)
The power dissipated is simply V^2/R
where V = 120 volts RMS
and R = 60 Ω
Answer:
α = 5 10⁻³ rad / s²
Explanation:
For this exercise we can use Newton's second law for rotational movement, where the force is electric
τ = I α
Where the torque is
τ = F x r = F r sin θ
Strength is
F = q E
The moment of inertia of a small ball, which we approximate to a point is
I = m r²
We replace
2 (q E) r sin θ = 2m r² α
The number 2 is because the two forces create the same torque
α = q E sin θ
/ m r
Let's reduce the magnitudes to the SI system
m = 1.0g = 1.0 10⁻³ kg
L = 2.0 cm = 2.0 10⁻² m
q = 10 nc = 10 10⁻⁹ C
E = 1.0 10 N / C
r = L / 2
r = 1.0 10⁻² m
Let's calculate
α = 10 10⁻⁹ 1.0 10 sin 30 / 1.0 10⁻³ 1.0 10⁻²
α = 5 10⁻³ rad / s²
