A 200 g block hangs from a spring with spring constant 10 N/m. At t = 0 s the block is 20 cm below the equilibrium point and moving upward with a speed of 100 cm/s. What are the block's

a. Oscillation frequency?

Question

A 200 g block hangs from a spring with spring constant 10 N/m. At t = 0 s the block is 20 cm below the equilibrium point and moving upward with a speed of 100 cm/s. What are the block's

a. Oscillation frequency?

Accepted Answer

Hey, everyone in this problem, we're asked to consider an ideal spring with a spring content of 15 newtons per meter that is suspended from a rigid support. A load of 350 g is attached to the springs free end at time. T A motion detector shows that the load is located 15 centimeters under the equilibrium position and is moving straight up at 0.5 m per second. We're asked to calculate the frequency of the moving load. We're given four answer traces, option A 0.5 Hertz, option B one Hertz, option C 1.5 Hertz and option D five Hertz. Now we're asked to calculate a frequency and we want to think about this. This is vertical, this spring is hanging. So this vertical simple harmonic motion where this um load is gonna be oscillating. What we wanna do is recall that we can relate the period to this spring constant through the following. The period is equal to two pi multiplied by the square root of M divided by K. Now we're looking for frequency, we're not looking for the period. But we can also recall that the period is equal to one divided by the frequency. So instead of having the period here, we can replace it with one divided by F. So the left hand side of our equation is now one divided by F that is equal to two pi multiplied by the square root of M divided by K. And we can write this as F is equal to one divided by two pi multiplied by the square root of K divided by M. OK. So we've just taken one over the right hand side of that equation. Oh What this means is that in order to find that frequency, all we need is a mass and the spring constant. And we don't even need to use this additional information about the time and the speed and all of that. OK. All right. So what is K K is going to be equal to noons per meter K in the spring constant? And the mass M we're given is 350 g. We wanted to convert this to our standard unit of kilogram. So we multiply in every one kg, there are 1000 g. So we multiply by one kg divided by 1000 g. The unit of gram divides out what we're doing is essentially dividing by to convert grams to kilograms. And we get that this is equal to 0.35 kg. All right. So now we can substitute these values into our equation. We have that the frequency is equal to one divided by two pi multiplied by the square root, the 15 newtons per meter divided by 0. kilograms. Now we're called that a newton is a kilogram meter per second squared. So we have kilogram meter per second squared divided by meter divided by kilogram. So inside of this square root, we're left with per second squared in terms of our unit. When we take the square root, it'll just be per second, which is equivalent to a Hertz, which is what we want when we're talking about units for frequency. And so the units work out here, which is always what we want. Now, if you work this out on your calculator, you're gonna find a frequency that is equal to 1.42 Hertz. Now, before we look at our answer choices, I wanna make one note. OK. This equation that we started with with our period, this is the same as when we're talking about horizontal simple harmonic motion. OK. So all that changes by having the effect of gravity on that vertical simple harmonic motion is the actual equilibrium position of the object. OK? Because gravity is going to have some effect on where that equilibrium position is by pulling it down a bit. But the actual period, the equation is the same. All right. So we found our frequency with 1.42 Hertz if we round this to two significant digits. We see that this corresponds with answer choice B one Hertz. Thanks everyone for watching. I hope this video helped see you in the next one.

A 200 g block hangs from a spring with spring constant 10 N/m. At t = 0 s the block is 20 cm below the equilibrium point and moving upward with a speed of 100 cm/s. What are the block's a. Oscillation frequency?

Verified video answer for a similar problem:

Key Concepts

Spring Constant

Mass-Spring System

Oscillation Frequency