Which of the following does NOT describe the conversion of potential to kinetic energy?

Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri

igor_vitrenko [27]

Answer:

a) the particles are 0.217 m apart

b) the particles are moving in the same direction.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

= -0.155 A + 0.372 A

= 0.217 A

since A = 1 m

Thus,

Δx = 0.217 m

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

= (-πA / T) sin(2πt / T)

= -(π(1) / 1.5) sin(2π(0.45) / 1.5)

= -1.99

and,

v₂ = dx₂ / dt

= (-πA / T) sin((2πt / T) + π/6)

= -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

= -1.40

Since both v₁ and v₂ are negative, this shows that the particles are moving in the same direction.

6 0

3 years ago

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c

IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

Explanation: 

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

$\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}$