Simple harmonic motion spring question... finding the period of oscillation!?
This one's a 2-parter... however, I did get the first part. A spring is suspended from a ceiling. When a 1.19kg mass is hung from its lower end, it stretches by 10.3 cm. (a) what is the spring constant? -- 113N/m (b) the mass is now displaced by an additional 4.9 cm and is released from rest. Find the ensuing period of oscillation. I tried finding a new spring constant (which incorporates the 4.9cm), then plugging it into T = 2*pi*(m/k)^1/2, but that doesn't give me the right answer... how do you approach this question? Thanks!
During the swing of a frictionless pendulum, what energy form(s) remain constant?
the answer is actually E. the reason is this. If a pendulum swings without friction it is experiencing a physics phenomena known as simple harmonic motion(SHM) during shm, the pendulum swings continuosly without stopping from left to right as there is no air friction. when the pendulum is at the extreme positions of left and right u notice that the speed there is the slowest because it is about ti change direction jus like when u throw a ball into the air and then it starts to fall at the max pt. Since KE=1/2MV^2, at the extremes of left and right, KE is zero. By conservation of energy, KE loss=PE gain, therefore 0 KE=max PE however when the pendulum is in the middle of oscillation at about the centre plane. (I.E: the centre point of oscillation) its velocity is the greatest, when V is greatest, so is KE. thus by COE, when KE is maximum, PE is minimum. hence KE and PE are constantly changing and are not constant throughout. If u wanna be sure try, it yourself in the lab. you'll notice the pendulum moves slowest at extreme positions and fastest in the middle. Therefore when u play a shooting game when the target moves left and right, the best time to shoot the target will be at its extreme positions as it moves slowest there(FYI) damn that was long, hope it helped u understand better anyway!! :)
Why is the formula of simple harmonic motion of pendulum, T=2π√(l/g) only apply to small angle (I have seen many explanation online but still don't get it)? What does the approximation sinθ≈θ have to do with the derivation of the pendulum formula?
Simple harmonic motion requires that the restoring force, that tends to make the object return to the equilibrium (center) position should be linear. In other words it should be directly proportional to the distance from the center position. If you were to graph the force against the distance, it should be a straight line. If it’s not, then it’s not “simple”, and you’d need a more complex formula than the one you’re using.In a pendulum, the restoring force is gravity, which always pulls straight down. But the actual restoring force is only a portion of the weight of the pendulum bob. At the bottom of the arc of travel, the restoring force is zero, because gravity does not push the bob sideways. The horizontal component of the weight is always zero.But lift the bob to a full 90°, and the restoring force is the full weight of the bob. And that’s fine, but the problem is that as you move from 0° to 90°, the graph is not a straight line. In fact, it follows the graph of a sine curve. So technically, it is not simple harmonic motion.But wait, we can still use the formula above if we are willing to keep the angle of the swing small. The reason is that, around the value of 0°, the graph of a sine curve is pretty straight. And the closer to zero you keep the angle, the more closely the curve approaches a straight line.The mathematical way of saying that it’s fairly linear is to say that [math]\sin \theta[/math] is approximately equal to [math]\theta[/math], measured in radians, to some acceptable approximation.How small is small? Well, that depends on how much precision you desire in your answer. How acceptable is the approximation? The more precise an answer you require, the smaller the angle has to be.
Why is the formula of simple harmonic motion of pendulum, T=2π√(l/g) only apply to small angle (I have seen many explanation online but still don't get it)? What does the approximation sinθ≈θ have to do with the derivation of the pendulum formula?
Simple harmonic motion requires that the restoring force, that tends to make the object return to the equilibrium (center) position should be linear. In other words it should be directly proportional to the distance from the center position. If you were to graph the force against the distance, it should be a straight line. If it’s not, then it’s not “simple”, and you’d need a more complex formula than the one you’re using.In a pendulum, the restoring force is gravity, which always pulls straight down. But the actual restoring force is only a portion of the weight of the pendulum bob. At the bottom of the arc of travel, the restoring force is zero, because gravity does not push the bob sideways. The horizontal component of the weight is always zero.But lift the bob to a full 90°, and the restoring force is the full weight of the bob. And that’s fine, but the problem is that as you move from 0° to 90°, the graph is not a straight line. In fact, it follows the graph of a sine curve. So technically, it is not simple harmonic motion.But wait, we can still use the formula above if we are willing to keep the angle of the swing small. The reason is that, around the value of 0°, the graph of a sine curve is pretty straight. And the closer to zero you keep the angle, the more closely the curve approaches a straight line.The mathematical way of saying that it’s fairly linear is to say that [math]\sin \theta[/math] is approximately equal to [math]\theta[/math], measured in radians, to some acceptable approximation.How small is small? Well, that depends on how much precision you desire in your answer. How acceptable is the approximation? The more precise an answer you require, the smaller the angle has to be.
Is angular velocity constant in simple harmonic motion?
I completely understand your confusion and was confused by this fact for a very long time. The confusion comes from the fact that in harmonic and rotational motion the letter [math]\omega[/math] means two different things and both are measured in radians per second.The first thing omega stands for and probably the first you learn of in a physics class is angular velocity, the rate of change of an angle.[math]\omega=\dfrac{\Delta \theta}{\Delta t}[/math]The second thing omega stands for is angular frequency. Good ol’ everyday frequency [math]f[/math] measures cycles per second so is calculated by dividing one cycle by how long that cycle took (the period).[math]f=\dfrac{1}{T}[/math]Angular frequency measures the same thing but instead of cycles it uses radians. Looking at a graph of the trig functions (no coefficients in front of [math]x[/math]) it can be seen that one cycle is [math]2\pi[/math] radians.Now we have that [math]\omega=2\pi f=\dfrac{2\pi}{T}[/math].(This can be checked by looking at the dimensions. [math]f[/math] is cycles per second and there are [math]2\pi[/math] radians per cycle. Multiplying cycles per second by radians per cycle gives radians per second which is what angular frequency is measured in.)Let’s look at an example: a pendulum. Pendulums swing back and forth with simple harmonic motion. Since the period for a pendulum is constant, the angular frequency is constant too. However, as the pendulum swings back and forth, [math]\frac{\Delta \theta}{\Delta t}[/math] is always changing with time! At the bottom of the swing the bob swings the fastest but at the top the bob has stopped to turn around.(In the world of calculus, to distinguish between angular velocity and frequency, the rate of change of the angle is represented by putting a dot on top of the angle: [math]\dot{\theta}[/math]. Now instead of two omegas there is one omega for angular frequency and [math]\dot{\theta}[/math] for angular velocity)I hope this helped clear up your confusion!
Is angular velocity constant in simple harmonic motion?
I completely understand your confusion and was confused by this fact for a very long time. The confusion comes from the fact that in harmonic and rotational motion the letter [math]\omega[/math] means two different things and both are measured in radians per second.The first thing omega stands for and probably the first you learn of in a physics class is angular velocity, the rate of change of an angle.[math]\omega=\dfrac{\Delta \theta}{\Delta t}[/math]The second thing omega stands for is angular frequency. Good ol’ everyday frequency [math]f[/math] measures cycles per second so is calculated by dividing one cycle by how long that cycle took (the period).[math]f=\dfrac{1}{T}[/math]Angular frequency measures the same thing but instead of cycles it uses radians. Looking at a graph of the trig functions (no coefficients in front of [math]x[/math]) it can be seen that one cycle is [math]2\pi[/math] radians.Now we have that [math]\omega=2\pi f=\dfrac{2\pi}{T}[/math].(This can be checked by looking at the dimensions. [math]f[/math] is cycles per second and there are [math]2\pi[/math] radians per cycle. Multiplying cycles per second by radians per cycle gives radians per second which is what angular frequency is measured in.)Let’s look at an example: a pendulum. Pendulums swing back and forth with simple harmonic motion. Since the period for a pendulum is constant, the angular frequency is constant too. However, as the pendulum swings back and forth, [math]\frac{\Delta \theta}{\Delta t}[/math] is always changing with time! At the bottom of the swing the bob swings the fastest but at the top the bob has stopped to turn around.(In the world of calculus, to distinguish between angular velocity and frequency, the rate of change of the angle is represented by putting a dot on top of the angle: [math]\dot{\theta}[/math]. Now instead of two omegas there is one omega for angular frequency and [math]\dot{\theta}[/math] for angular velocity)I hope this helped clear up your confusion!
What are the functions and purposes of a simple pendulum?
One of the more fascinating uses of the pendulum is Foucault’s pendulum. A Foucault pendulum demonstrates that the Earth rotates.(I think) Foucault’s pendulum proves that the fixed (non-rotating) Earth is not an inertial frame of reference, because the rotation of the Earth can be measured without reference to the outside (e.g. following the night time path of the stars or the rising and setting Sun).The picture below is a Foucault’s pendulum at Carleton University in Ottawa Canada.The long (several stories high) pendulum swings back and forth and traces around the map of the world on the floor once per day.
What are the functions and purposes of a simple pendulum?
One of the more fascinating uses of the pendulum is Foucault’s pendulum. A Foucault pendulum demonstrates that the Earth rotates.(I think) Foucault’s pendulum proves that the fixed (non-rotating) Earth is not an inertial frame of reference, because the rotation of the Earth can be measured without reference to the outside (e.g. following the night time path of the stars or the rising and setting Sun).The picture below is a Foucault’s pendulum at Carleton University in Ottawa Canada.The long (several stories high) pendulum swings back and forth and traces around the map of the world on the floor once per day.
Simple harmonic motion?
your basic equation is y = 0.204 sin (a t) where a[2π] = 2.80--> a = 1.40 / π y ' {max} = 0.204 (1.40/π) ≈ 1/11 m / s
How to calculate the period of the swing of simple pendulum?
Let’s do the general theory first and feed in numbers later. Imagine a particle P of mass m, attached to a light string of length r, fixed at one end O. Let OA be vertical and let the angle made by the string to the vertical at any time t be θ. The angular acceleration of P is d²θ/dt² away from OA (the sense in which θ increases). P also has an acceleration r(dθ/dt)² towards O (centripetal acceleration). Applying Newton’s 2nd Law along the tangent (F = Ma): mgsinθ = -mr d²θ/dt², and since θ is always very small sinθ ≈ θ Hence mgθ ≈ -mr d²θ/dt² ⇒ d²θ/dt² = -(g/r)θ This is the basic equation of angular Simple Harmonic Motion. So a particle oscillating through small angles at the end of a light string performs angular SHM with a period of oscillation T given by T = 2π/n where n² = g/r ∴ Period T = 2π√(r/g) So in your problem Period T = 2π√(0.82/9.81) = 1.81657….seconds ≈ 1.817 seconds to 3dp