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Tom [10]
3 years ago
11

A particle of mass m moves under the influence of a force given by F = (−kx + kx3/α2) where k and α are positive constants. a) F

ind the potential energy function U(x) (Take U(x=0) = 0). b) Find any equilibrium points and determine if they are stable or unstable. c) Sketch the potential as a function of position. d)What is the maximum total energy, Emax a particle may have if its motion is to remain bounded? e) For the case of bounded motion with E
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
4 0

Answer:

Explanation:

a ) F = (-kx + kx³/a²)

intensity of field

I = F / m

=  (-kx + kx³/a²) / m

If U be potential function

- dU / dx =  (-kx + kx³/a²) / m

U(x)  = ∫  (kx - kx³/a²) / m dx

= k/m ( x²/2 - x⁴/4a²)

b )

For equilibrium points , U is either maximum or minimum .

dU / dx = x - 4x³/4a² = 0

x = ± a.

dU / dx = x - x³/a²

Again differentiating

d²U / dx² = 1 - 3x² / a²

Put the value of x = ± a.

we get

d²U / dx²  = -2 ( negative )

So at x = ± a , potential energy U is maximum.

c )

U =  k/m ( x²/2 - x⁴/4a²)

When x =0 , U = 0

When x=  ± a.

U is maximum

So the shape of the U-x curve is like a bowl centered at x = 0

d ) Maximum potential energy

put x = a or -a in

U(max)  =  k/m ( x²/2 - x⁴/4a²)

= k/m ( a² / 2 - a⁴/4a²)

= k/m ( a² / 2 - a²/4)

a²k / 4m

This is the maximum total energy where kinetic energy is zero.

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In most cases, what happens to a liquid when it cools?
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Option (A) and (F)

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5 0
3 years ago
Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t
adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x  (1)

Also, the T₂ = t/n₂ = t/(n₁ + x)  (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

6 0
3 years ago
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