SMH-Block Attached to a spring

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In summary: You are right.In summary, a block with mass 1.7 kg attached to a spring with a period of 4.3 s and amplitude of 29.6 cm undergoes simple harmonic motion on a frictionless surface. When the spring is unstretched, the block is 1.7 m away from a wall. When the spring is snapped at the equilibrium point, the block moves towards the wall with a constant velocity of Aω, and it takes time 't' to reach the wall, where 't' can be calculated using the equation s = ut.
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Physicist1234
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Homework Statement


A block with mass m = 1.7 kg attached to the end of a spring undergoes simple harmonic motion on a horizontal frictionless surface. The oscillation period is T = 4.3 s and the oscillation amplitude is A = 29.6 cm. When the spring is unstretched the block sits at a distance L = 1.7 m from a wall. Suppose we snap the spring at the very moment the block passes through the equilibrium point towards the wall. Calculate the time it takes for the block to hit the wall.

Homework Equations


w=2*Pi*f=2*Pi/T=sqrt(k/m)
x=Acos(wt)

The Attempt at a Solution


w=2*Pi/4.3=1.46. I'm not sure how to go about using the above formulas
 
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  • #2
Please use proper signs and symbols. By hitting the 'Σ' button, you see vast signs and symbols. Use subscript and superscript. Use 'x' for into instead of *.

Its too hard to understand what equations you typed. Please follow above tips and edit your thread.
 
  • #3
Physicist1234 said:

Homework Statement


A block with mass m = 1.7 kg attached to the end of a spring undergoes simple harmonic motion on a horizontal frictionless surface. The oscillation period is T = 4.3 s and the oscillation amplitude is A = 29.6 cm. When the spring is unstretched the block sits at a distance L = 1.7 m from a wall. Suppose we snap the spring at the very moment the block passes through the equilibrium point towards the wall. Calculate the time it takes for the block to hit the wall.

Homework Equations


w=2*Pi*f=2*Pi/T=sqrt(k/m)
x=Acos(wt)

The Attempt at a Solution


w=2*Pi/4.3=1.46. I'm not sure how to go about using the above formulas
Calculate the velocity when the block is at the point when the spring is just relaxed. You cut the spring, so the block moves with that velocity towards the wall, 1.7 m away. How long does it take to reach the wall?
 
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  • #5
Else, since spring was snapped, there won't be any acceleration by any source. Means block moves with constant velocity, taking velocity at mean position Aω as initial and final as 0 (since block crashes the wall) Apply s = ut (s=1.7m, u=Aω2), you get required value 't'.
 
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  • #6
AlphaLearner said:
Else, since spring was snapped, there won't be any acceleration by any source. Means block moves with constant velocity, taking velocity at mean position Aω2 as initial and final as 0 (since block crashes the wall) Apply s = ut (s=1.7m, u=Aω2), you get required value 't'.
The speed at the equilibrium position is Aω, and constant after the spring is cut.
 
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  • #7
ehild said:
The speed at the equilibrium position is Aω, and constant after the spring is cut.
True, it is Aω
V = ω√A2-x2
At mean position 'x' is 0. By solving, we get Aω.
Sorry for wrong information.
 

1. What is an SMH-Block attached to a spring?

An SMH-Block attached to a spring is a simple harmonic motion (SMH) system consisting of a mass (block) attached to a spring. The mass is able to move back and forth in a repetitive motion due to the restoring force of the spring.

2. What is the equation for the motion of an SMH-Block attached to a spring?

The equation for the motion of an SMH-Block attached to a spring is x(t) = A*cos(ωt + φ), where x is the displacement of the mass, A is the amplitude of the motion, ω is the angular frequency, and φ is the phase angle.

3. What is the relationship between the mass, spring constant, and frequency in an SMH-Block attached to a spring?

The mass, spring constant, and frequency are all related in an SMH-Block attached to a spring. The frequency (f) is equal to the square root of the spring constant (k) divided by the mass (m): f = √(k/m). This means that increasing the mass or decreasing the spring constant will result in a lower frequency and slower motion, while decreasing the mass or increasing the spring constant will result in a higher frequency and faster motion.

4. How does the amplitude affect the motion of an SMH-Block attached to a spring?

The amplitude of an SMH-Block attached to a spring determines the maximum displacement of the mass and therefore affects the energy and speed of the motion. A larger amplitude will result in a higher energy and faster motion, while a smaller amplitude will result in a lower energy and slower motion.

5. What factors can affect the motion of an SMH-Block attached to a spring?

The motion of an SMH-Block attached to a spring can be affected by several factors, including the mass of the block, the stiffness of the spring, the amplitude of the motion, and any external forces acting on the system. Other factors such as friction and air resistance can also play a role in the motion of the system.

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