Solving an Oscillating Block's Period and Speed

[prophet05]

Homework Statement

A 500g block is attached to a spring on a frictionless horizontal surface. The block is pulled to stretch the spring by 10cm, then gently released. A short time later, as the block passes through the equilibrium position, its velocity is 1m/s.
>A)What is the block's period of oscillation?
>B)What is the block's speed at the point where the spring is compressed by 5cm?

Homework Equations

T = 2(pi)sqrt(I/mgd)

The Attempt at a Solution

I'm having troubles starting this problem. I'm thinking since it's 1m/s at it's equilibrium position (5cm) then it take 20seconds for half a cycle? That sounds completely wrong. I need some help.

 

[denverdoc]

Homework Statement

A 500g block is attached to a spring on a frictionless horizontal surface. The block is pulled to stretch the spring by 10cm, then gently released. A short time later, as the block passes through the equilibrium position, its velocity is 1m/s.
>A)What is the block's period of oscillation?
>B)What is the block's speed at the point where the spring is compressed by 5cm?

Homework Equations

T = 2(pi)sqrt(I/mgd)

The Attempt at a Solution

I'm having troubles starting this problem. I'm thinking since it's 1m/s at it's equilibrium position (5cm) then it take 20seconds for half a cycle? That sounds completely wrong. I need some help.

Have you encountered an equation that looks like x=Asin(w*t) ??

This is a general eqn for describing simple harmonic motion, which is what this problem is about. It also describes the motion of a pendulum.

w*t (angular velocity * time) in the sine expression above, is the key to answering this problem as it also determines the period thru the relation,

T=2*pi/w.

So how to determine w?

Well if we were given a value of time and position we could do so as we are given A, the amplitude, as equal to 10cm.

But all we are told is that a "short time later", its velocity is 1m/s as it passes thru the equalibrium point. That is when x=0.

We can differentiate the above expression with respect to time to get,

dx/dt=v=w*A*cos(wt). Since we know that x=0, it follows sin(wt)=0 at that time, and most importantly for the purposes of this problem cos(wt)=1 at the same time, so we can substitute 1 for cos(wt).

Can you finish from here?

 

[m2003]

I would approach this problem by first identifying the relevant equations and variables. In this case, the period of oscillation (T) can be calculated using the equation T = 2(pi)sqrt(I/mgd), where I is the moment of inertia of the block, m is its mass, g is the acceleration due to gravity, and d is the displacement from the equilibrium position.

To calculate the period, we first need to find the moment of inertia of the block. Since the block is attached to a spring on a frictionless surface, we can assume that it will undergo simple harmonic motion. In this case, the moment of inertia can be calculated as I = md^2, where d is the distance from the center of mass to the axis of rotation. In this problem, the block is moving horizontally, so we can assume that its center of mass is at the center of the block and d = 0. Therefore, the moment of inertia is simply I = 0.

Now, we can substitute the values into the equation for T: T = 2(pi)sqrt(0/mgd) = 0. This means that the period of oscillation is 0 seconds, which is not possible. This suggests that there may be an error in the given information or the problem statement.

For part B, we can use the equation for speed, v = sqrt(k/m)x, where k is the spring constant and x is the displacement from the equilibrium position. We know that the spring is compressed by 5cm (0.05m), so we can plug in the values to solve for v: v = sqrt(k/0.5)(0.05) = sqrt(2k). However, we do not have enough information to solve for k, so we cannot determine the block's speed at this point.

In conclusion, there may be an error in the given information or the problem statement. I would recommend double-checking the given values and making sure that all necessary information is provided in order to accurately solve this problem.