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A particle of mass m moves in one dimensional pote
ntial v(x)=-ax2+ bx4where a and b are positive constants. What is the angular frequency of small oscillationsabout the minimum of the potential in units of root a/m.
Correct answer is '2'. Can you explain this answer?
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A particle of mass m moves in one dimensional pote... morential v(x)=-...
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Given Information:
A particle of mass m moves in a one-dimensional potential given by v(x) = -ax^2 + bx^4, where a and b are positive constants.
To find:
The angular frequency of small oscillations about the minimum of the potential in units of √(a/m).
Solution:
Step 1: Finding the Minimum of the Potential
To find the minimum of the potential, we need to find the value of x for which the derivative of v(x) with respect to x is equal to zero.
dv(x)/dx = -2ax + 4bx^3
Setting the derivative equal to zero and solving for x:
-2ax + 4bx^3 = 0
2ax = 4bx^3
x(2a - 4bx^2) = 0
x = 0 or x^2 = a/2b
Since a and b are positive constants, the minimum of the potential occurs at x = 0.
Step 2: Taylor Expansion of the Potential
To determine the angular frequency of small oscillations, we can use the Taylor expansion of the potential around the minimum (x = 0).
v(x) = v(0) + (dv/dx)|x=0 * x + (1/2)(d^2v/dx^2)|x=0 * x^2 + ...
The first two terms of the Taylor expansion are sufficient for our calculation.
v(x) = v(0) + (1/2)(d^2v/dx^2)|x=0 * x^2
Since the potential is minimized at x = 0, v(0) = 0.
v(x) = (1/2)(d^2v/dx^2)|x=0 * x^2
Since the potential is given by v(x) = -ax^2 + bx^4, we can find the second derivative:
d^2v/dx^2 = -2a + 12bx^2
Evaluating the second derivative at x = 0:
(d^2v/dx^2)|x=0 = -2a
Step 3: Harmonic Approximation
Using the harmonic approximation, we can write the potential as:
v(x) = (1/2)(d^2v/dx^2)|x=0 * x^2
v(x) = -a * x^2
This is the equation of a simple harmonic oscillator with angular frequency ω = √(a/m).
Step 4: Finding the Angular Frequency in Units of √(a/m)
The angular frequency of the small oscillations is ω = √(a/m).
Since a and m are positive constants, the angular frequency in units of √(a/m) is given by:
ω/(√(a/m)) = √(a/m) / √(a/m) = 1
Therefore, the angular frequency of small oscillations about the minimum of the potential in units of √(a/m) is 2.
A particle of mass m moves in a one-dimensional potential given by v(x) = -ax^2 + bx^4, where a and b are positive constants.
To find:
The angular frequency of small oscillations about the minimum of the potential in units of √(a/m).
Solution:
Step 1: Finding the Minimum of the Potential
To find the minimum of the potential, we need to find the value of x for which the derivative of v(x) with respect to x is equal to zero.
dv(x)/dx = -2ax + 4bx^3
Setting the derivative equal to zero and solving for x:
-2ax + 4bx^3 = 0
2ax = 4bx^3
x(2a - 4bx^2) = 0
x = 0 or x^2 = a/2b
Since a and b are positive constants, the minimum of the potential occurs at x = 0.
Step 2: Taylor Expansion of the Potential
To determine the angular frequency of small oscillations, we can use the Taylor expansion of the potential around the minimum (x = 0).
v(x) = v(0) + (dv/dx)|x=0 * x + (1/2)(d^2v/dx^2)|x=0 * x^2 + ...
The first two terms of the Taylor expansion are sufficient for our calculation.
v(x) = v(0) + (1/2)(d^2v/dx^2)|x=0 * x^2
Since the potential is minimized at x = 0, v(0) = 0.
v(x) = (1/2)(d^2v/dx^2)|x=0 * x^2
Since the potential is given by v(x) = -ax^2 + bx^4, we can find the second derivative:
d^2v/dx^2 = -2a + 12bx^2
Evaluating the second derivative at x = 0:
(d^2v/dx^2)|x=0 = -2a
Step 3: Harmonic Approximation
Using the harmonic approximation, we can write the potential as:
v(x) = (1/2)(d^2v/dx^2)|x=0 * x^2
v(x) = -a * x^2
This is the equation of a simple harmonic oscillator with angular frequency ω = √(a/m).
Step 4: Finding the Angular Frequency in Units of √(a/m)
The angular frequency of the small oscillations is ω = √(a/m).
Since a and m are positive constants, the angular frequency in units of √(a/m) is given by:
ω/(√(a/m)) = √(a/m) / √(a/m) = 1
Therefore, the angular frequency of small oscillations about the minimum of the potential in units of √(a/m) is 2.
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A particle of mass m moves in one dimensional pote... morential v(x)=-ax2+ bx4where a and b are positive constants. What is the angular frequency of small oscillationsabout the minimum of the potential in units of root a/m.Correct answer is '2'. Can you explain this answer?
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