Answer:
Explanation:
Given a particle of mass
M = 1.7 × 10^-3 kg
Given a potential as a function of x
U(x) = -17 J Cos[x/0.35 m]
U(x) = -17 Cos(x/0.35)
Angular frequency at x = 0
Let find the force at x = 0
F = dU/dx
F = -17 × -Sin(x/0.35) / 0.35
F = 48.57 Sin(x/0.35)
At x = 0
Sin(0) =0
Then,
F = 0 N
So, from hooke's law
F = -kx
Then,
0 = -kx
This shows that k = 0
Then, angular frequency can be calculated using
ω = √(k/m)
So, since k = 0 at x = 0
Then,
ω = √0/m
ω = √0
ω = 0 rad/s
So, the angular frequency is 0 rad/s
Answer:
a
Explanation:
water is only able to desolve two substances , salt , sugar,
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is

Which means that the frequency is

and the angular frequency is

In a spring-mass system, the maximum velocity of the object is given by

where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
On Earth, the acceleration of gravity is 9.8 m/s² downward.
So any object with only gravity acting on it gains 9.8 m/s of
downward speed every second.
If the rock starts out moving upward at 10 m/s, then it will
continue upward for only (10/9.8) = 1.02 second, before
it stops rising and starts falling.
Its average speed during that time is (1/2) (10 + 0) = 5 m/s .
At an average speed of 5 m/s for 1.02 sec,
the rock rises
(5 m/s) x (1.02 sec) = 5.102 meters .