How Long for a Block in SHM to Hit a Wall After Spring Snaps?

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A block with a mass of 1.7 kg attached to a spring undergoes simple harmonic motion, with a period of 4.3 seconds and an amplitude of 29.6 cm. When the spring is snapped while the block is at the equilibrium point, it moves towards a wall located 1.7 m away. The velocity of the block at this point is calculated as Aω, where ω is derived from the period. After the spring is cut, the block travels with constant velocity until it hits the wall. The discussion emphasizes using the formula s = ut to determine the time taken to reach the wall.
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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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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.
 
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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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 as initial and final as 0 (since block crashes the wall) Apply s = ut (s=1.7m, u=Aω2), you get required value 't'.
 
Last edited:
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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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.
 
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