# Horizontal Harmonic motion and Springs

• mmiller39
In summary, the question asks for the initial velocity needed for a block attached to a spring to knock over a bottle located 0.080 m to the right of its unstrained position. With an angular frequency of 9.5 rad/s and a spring stretching of 0.050 m, the block must have a minimum amplitude of 0.080 m to achieve this. The initial velocity needed can be found by considering the energy the block needs to reach this amplitude.
mmiller39
Hi all,

I am having difficulty with a question. I would like some guidance.

QUESTION:

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 9.5 rad/s. The drawing shows the position of the block when the spring is unstrained. This position is labeled ''x = 0 m.'' The drawing also shows a small bottle located 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by 0.050 m, and is then thrown to the left. In order for the block to knock over the bottle,it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.

http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c10/ch10p_34.gif

http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c10/ch10p_34.gif

MY WORK:

What we know

A1 = .050 m
A2 = .080 m
f = 1.51197 <----- 9.5 = 2pi f solve for f

at this point I don't know how to incorporate an initial velocity. Any help would be greatly appreciated.

Thanks.

Last edited:
It's not that there are two amplitudes in the problem. There is one amplitude that must be at least .08m and to get that amplitude the block starts from a certain position x_o = .05m with some initial velocity v_o. Think about the energy the block must have to achieve the needed amplitude.

Hello,

Thank you for reaching out for guidance with this question. It seems like you have a good understanding of the basic principles of simple harmonic motion and the given parameters in this scenario. To incorporate the initial velocity, we can use the equation for the maximum velocity in simple harmonic motion, which is v_max = w*A, where w is the angular frequency and A is the amplitude (maximum displacement from equilibrium). In this case, we can use A = 0.050 m as the block is pulled to the right by this distance before being thrown to the left. Therefore, the maximum velocity at this point would be v_max = (9.5 rad/s)*(0.050 m) = 0.475 m/s.

Now, to knock over the bottle, the block must have enough speed to reach the bottle's location, which is 0.080 m to the right of the equilibrium position. This means that the block must have a velocity of at least 0.080 m/s when it reaches this position. Since the block is initially moving to the left, this means that it must have a minimum initial velocity of v0 = 0.080 m/s + 0.475 m/s = 0.555 m/s in order to knock over the bottle. I hope this helps and let me know if you have any further questions. Good luck with your work!

## 1. What is horizontal harmonic motion?

Horizontal harmonic motion is a type of motion in which an object moves back and forth in a straight line with a constant amplitude and frequency. This type of motion can be observed in systems such as springs, pendulums, and vibrating strings.

## 2. How is horizontal harmonic motion related to springs?

Springs are commonly used to study horizontal harmonic motion because they exhibit a restoring force that is proportional to the displacement from equilibrium. This means that as the spring is stretched or compressed, it exerts a force in the opposite direction, causing the object attached to it to oscillate back and forth.

## 3. What is the equation for horizontal harmonic motion?

The equation for horizontal harmonic motion is x = A sin(ωt + φ), where x is the displacement from equilibrium, A is the amplitude or maximum displacement, ω is the angular frequency, and φ is the phase angle. This equation describes the position of the object at any given time during the motion.

## 4. How do you calculate the period of horizontal harmonic motion?

The period of horizontal harmonic motion is the time it takes for one complete cycle of the motion. It can be calculated using the equation T = 2π/ω, where T is the period and ω is the angular frequency. This means that the period is dependent on the frequency of the motion.

## 5. What factors affect the amplitude of horizontal harmonic motion?

The amplitude of horizontal harmonic motion is affected by the initial displacement, the mass of the object, and the spring constant. A larger initial displacement or a higher spring constant will result in a larger amplitude, while a heavier mass will result in a smaller amplitude.

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