A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 2.0 m/s. Determine the amplitude A of the motion.
x = A sin 2 pi f t
v = 2 pi f A cos 2 pi f t
max speed is when cos = 1 or - 1
vmax = 2 pi f A = 2
F = m a = -kx
a = -(2 pi f)^2 A sin 2 pi f t
= - (2 pi f)^2 x
so
.2 (-2pi f)^2 x = -120 x
so
2 pi f = sqrt (k/m) but you knew that :)
2 pi f = sqrt(120/.2) = 24.5
but
2 pi f A = 2
24.5 A = 2
A = .0816 or about 8 centimeters
Well, Amplitude (A) is the maximum displacement from the equilibrium position. In this case, we can use the maximum speed (v_max) to find the amplitude.
Now, you're probably wondering how to relate speed to displacement in simple harmonic motion? Well, let me spring into action and explain!
In simple harmonic motion, the maximum speed occurs when the object is at the equilibrium position and all the potential energy stored in the spring is converted to kinetic energy.
The potential energy stored in the spring (PE_spring) is given by the equation PE_spring = (1/2) * k * A^2, where k is the spring constant and A is the amplitude.
And the kinetic energy (KE) of the object is given by the equation KE = (1/2) * m * v_max^2, where m is the mass of the object and v_max is the maximum speed.
Since the maximum speed occurs at the equilibrium position where the displacement is zero, the potential energy is also zero. Therefore, we can set PE_spring = 0.
Now we can equate the potential energy to the kinetic energy to express A in terms of v_max:
(1/2) * k * A^2 = (1/2) * m * v_max^2
Plugging in the given values, we have:
(1/2) * 120 N/m * A^2 = (1/2) * 0.20 kg * (2.0 m/s)^2
After some calculations, we can solve for the amplitude A. Hmm, let me do some math magic here... *abracadabra*
A = sqrt((0.20 kg * (2.0 m/s)^2) / 120 N/m)
*calculating, calculating...*
A ≈ 0.127 m
So, the amplitude of the motion is approximately 0.127 meters. Tada!
To determine the amplitude A of the motion, we can use the equation for the maximum speed of an object in simple harmonic motion:
v_max = ωA
Where v_max is the maximum speed, ω is the angular frequency, and A is the amplitude.
First, we need to find the angular frequency ω. The angular frequency is related to the spring constant k and the mass m of the object:
ω = √(k / m)
Given that the spring constant k is 120 N/m and the mass m is 0.20 kg, we can substitute these values into the equation to find ω:
ω = √(120 N/m / 0.20 kg)
ω = √600 rad/s
Now we can use the equation for v_max to find the amplitude A:
2.0 m/s = (√600 rad/s)A
Solving for A, we divide both sides of the equation by √600 rad/s:
A = 2.0 m/s / √600 rad/s
A ≈ 0.163 m (rounded to three decimal places)
Therefore, the amplitude A of the motion is approximately 0.163 m.
To determine the amplitude of the motion, we need to use the equation that relates the maximum speed of an object in simple harmonic motion to its amplitude.
The equation is:
v_max = ω * A
Where:
- v_max is the maximum speed of the object (given as 2.0 m/s)
- ω is the angular frequency of the motion (which depends on the spring constant and the mass of the object)
- A is the amplitude of the motion (what we're trying to find)
First, let's find the angular frequency using the equation:
ω = √(k / m)
Where:
- k is the spring constant (given as 120 N/m)
- m is the mass of the object (given as 0.20 kg)
Substituting the given values:
ω = √(120 N/m / 0.20 kg) = √(600 N/kg) ≈ 24.49 rad/s
Now we can rearrange the first equation and solve for the amplitude:
A = v_max / ω = 2.0 m/s / 24.49 rad/s = 0.0817 m
Therefore, the amplitude of the motion is approximately 0.0817 meters.