Considering a mass-spring system aligned horizontally moving at a velocity v.
a) How will the velocity of the mass change if amplitude is doubled.
b) How will the velocity of the mass change if the spring constant is quartered?
I know the answer to the first one is that it will be doubled but am not 100% sure if that is correct. I am stuck on the second question.
Thank you.
the energy stored in the spring is
... ½ k x²
... this energy is K.E. when the spring is in the neutral position (the middle of its travel)
K.E. = ½ m v²
... so the velocity and the amplitude are directly related to each other (both squared)
a) velocity will double (as you thought)
b) quartering the spring constant (k)
will quarter the stored energy
... since energy is related to the square of the velocity
... quartering k will halve the velocity ... √¼ = ½
a) To understand how the velocity of the mass in a mass-spring system changes when the amplitude is doubled, we need to refer to the equation of motion for simple harmonic motion.
The equation of motion for a mass-spring system is given by:
m * a = -k * x
where m is the mass of the object, a is the acceleration, k is the spring constant, and x is the displacement from the equilibrium position.
To relate this equation to the velocity of the mass, we can differentiate both sides of the equation with respect to time (t):
m * (d^2x/dt^2) = -k * x
Using the chain rule, we can express the acceleration (a) as the second derivative of the displacement (x) with respect to time (t).
Taking the derivative of the displacement equation with respect to time gives us:
m * d^2x/dt^2 = -k * x
Dividing both sides of the equation by m, we get:
d^2x/dt^2 = - (k/m) * x
Notice that the term -(k/m) represents the angular frequency (ω) of the simple harmonic motion:
ω = √(k/m)
The angular frequency is related to the velocity (v) of the mass as:
v = ω * √(A^2 - x^2)
where A is the amplitude of the simple harmonic motion.
Now, if we double the amplitude (A), the equation for velocity becomes:
v' = ω * √(4A^2 - x^2)
Since the velocity is directly proportional to the angular frequency, and the angular frequency remains constant for a given mass and spring constant, we can conclude that doubling the amplitude will also double the velocity.
b) To determine how the velocity of the mass changes if the spring constant (k) is quartered, we again refer to the equation of motion for a mass-spring system:
m * a = -k * x
As shown earlier, the angular frequency (ω) is given by:
ω = √(k/m)
Since the angular frequency is inversely proportional to the square root of the spring constant, we can derive the relationship between the velocity (v) and the spring constant (k):
v = ω * √(A^2 - x^2)
If we quarter the spring constant (k) by dividing it by 4, the new angular frequency (ω') can be expressed as:
ω' = √(¼k/m) = √(k'/m) [where k' is the new spring constant]
Since the mass (m) remains constant, we can conclude that quartering the spring constant will result in taking the square root of it. Therefore, the new angular frequency (ω') will be half of the original angular frequency (ω).
Using this relationship, we can determine that the new velocity (v') will be half of the original velocity (v).
In summary:
a) Doubling the amplitude of a mass-spring system will result in doubling the velocity of the mass.
b) Quartering the spring constant of a mass-spring system will result in halving the velocity of the mass.