An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.
What is the object's velocity when its potential energy is 2/3E?
the answer is
±√(k/m)(A/√3)
can you explain the √3 thank you
Certainly! In simple harmonic motion, the potential energy and kinetic energy of the system oscillate sinusoidally with time. The total mechanical energy (E) of the system remains constant and is the sum of the potential energy (PE) and kinetic energy (KE) at any given point in time.
Given that the potential energy is 2/3E, we can write the equation as follows:
PE = (2/3)E
Since the total mechanical energy (E) is the sum of the potential energy (PE) and kinetic energy (KE), we have:
E = PE + KE
By substituting the value of potential energy (PE) into the equation, we get:
E = (2/3)E + KE
Simplifying this equation, we find:
KE = E - (2/3)E
KE = (1/3)E
The kinetic energy (KE) is given by (1/2)mv^2, where m is the mass of the object and v is its velocity. Since KE = (1/3)E, we can equate the two equations:
(1/2)mv^2 = (1/3)E
Now, solving for v, we have:
v^2 = (2/3)(2E/m)
v^2 = (4/3)(E/m)
v = ± √[(4/3)(E/m)]
To simplify further, we can rewrite this expression as:
v = ± √[(4E/3) / m]
Simplifying the expression under the square root, we find:
v = ± √[(4/3)(E/m)]
Now, let's focus on the part within the square root, (4/3). We can rewrite it as (4/3) * (1/√1), which is (√3/√3) * (4/3). This gives us:
v = ± √[(4/3)(E/m)] * (√3/√3)
Multiplying the (√3/√3) with the expression under the square root, we have:
v = ± √[(4/3)(E/m)(√3/√3)]
Finally, simplifying the expression under the square root, we get:
v = ± √[(4/3)(E/m)(√3)]
Thus, the object's velocity when its potential energy is 2/3E is ±√(k/m)(A/√3).
Sure, I'd be happy to explain the factor of √3 in the given answer.
In simple harmonic motion, the total mechanical energy of the system (E) is the sum of the kinetic energy (K) and potential energy (U):
E = K + U
Given that the potential energy is 2/3E, we can write:
2/3E = U
Now, let's consider the potential energy in the spring. For an object attached to a spring undergoing simple harmonic motion, the potential energy is given by:
U = (1/2)kA^2
where k is the force constant of the spring and A is the amplitude (maximum displacement from equilibrium).
We can substitute this expression for U into the equation 2/3E = U:
2/3E = (1/2)kA^2
Simplifying this equation, we have:
(2/3E) / (1/2)k = A^2
So, A^2 = (4/3)(E/k).
Now, let's take the square root of both sides:
√A^2 = √[(4/3)(E/k)]
Since A is a positive quantity (displacement cannot be negative), we can write:
A = √[(4/3)(E/k)]
Now, let's consider the relationship between the maximum velocity (Vmax) and the amplitude (A) in simple harmonic motion. The maximum velocity occurs when the object passes through the equilibrium position. At this point, all the potential energy is converted into kinetic energy. The maximum kinetic energy (Kmax) is given by:
Kmax = (1/2)mVmax^2
Since the total mechanical energy is conserved, we can write:
E = Kmax + Umax
Since Umax = (1/2)kA^2, we have:
E = (1/2)mVmax^2 + (1/2)kA^2
Let's rearrange this equation to solve for Vmax:
Vmax^2 = [(2E - kA^2) / m]
Taking the square root of both sides:
Vmax = ±√[(2E - kA^2) / m]
Now, substitute the expression for A^2 we found earlier (A^2 = (4/3)(E/k)):
Vmax = ±√[(2E - (k/m)(4/3)E)]
Simplifying:
Vmax = ±√[(6E/3 - (4/3)(E)(k/m))]
Vmax = ±√[(2E/3)(3 - 2k/m)]
The factor of √3 comes from the expression (3 - 2k/m). When simplified, this term becomes √3. Therefore, the final answer for the velocity when the potential energy is 2/3E is:
Vmax = ±√[(2E/3)(√3)]
Vmax = ±√[(2/3)(E/√3)(√3)]
Vmax = ±√[(2/3)(k/m)(A/√3)]
Hence, the object's velocity when its potential energy is 2/3E is ±√[(k/m)(A/√3)].
let's say
x = A sin w t
v = A w cos w t
a = -A w^2 sin wt = -w^2 x
w = sqrt k/m
but we do not really need to know that
max potential energy = (1/2)k max x^2
= (1/2) k A^2
max kinetic energy = (1/2) m v^2
but max v^2 = A^2 w^2
so
max ke = (1/2) m A^2 w^2
the max ke = max pe
so
(1/2) k A^2 = (1/2) m A^2 w^2
so
w = sqrt(k/m) whew, that's settled
when it is stopped, all the energy is potential
so
E = (1/2) k A^2
(2/3) E = (1/3) k A^2 = potential energy
so where are we at then?
(1/2) k x^2 = (1/3) k A^2
x^2 = (2/3) A^2
x = (2/3)^.5 A
sin w t = (2/3)^.5
sin^2 wt = (2/3)
so
cos^2 w t = 1 - 2/3 = 1/3
remember v = A w cos w t
so
v^2 = A^2 w^2 cos^2 wt
v = +/- A w sqrt (1/3)
since w = sqrt (k/m)
we agree