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# Simple Harmonic Motion watch

1. "A mass of 0.40kg hangs from a spring. A hammer strikes it directly from below and imparts an impulse of 0.37Ns. How high above the equilibrium position does the mass travel if the natural angular frequency omega = 32?"

Could anyone please help me answer this? I've tried several times using various approaches but keep hitting a wall where I need the time take to the top. It's simple enough to find the initial velocity, but then to write v(t) out starts involving (omega^2)*(mass of object) = k[Spring Constant]. Please anybody help me on this equation... I am desperate.
2. (Original post by Afrikaans Boytjie)
"A mass of 0.40kg hangs from a spring. A hammer strikes it directly from below and imparts an impulse of 0.37Ns. How high above the equilibrium position does the mass travel if the natural angular frequency omega = 32?"

Could anyone please help me answer this? I've tried several times using various approaches but keep hitting a wall where I need the time take to the top. It's simple enough to find the initial velocity, but then to write v(t) out starts involving (omega^2)*(mass of object) = k[Spring Constant]. Please anybody help me on this equation... I am desperate.
Although probably it's possible to solve it considering times and so, I believe it's done easier by considering energy.

I've found two ways of dealing with this, one more simple, the other a bit longer. Both use conservation of energy, and for both you would need to express kinetic energy in terms of mass and initial momentum.

The simpler approach: the conservation of energy in harmonic motion says that at the maximum displacement all kinetic energy has changed into potential energy of SHM. Can you express total energy of SHM in terms of amplitude, mass and angular frequency?

The longer approach would not use SHM. You can consider transformations of energy changes during the motion: kinetic energy, elastic energy of the spring and gravitational potential energy. It's a much longer approach and requires much more work with transforming equations, but gives the same result in the end, and it may be very beneficial to see that for oneself.

Feel free to ask if you need more help.
3. Unless I'm missing something fundamental here..
The velocity v imparted by the hammer is the maximum velocity of the oscillations. That occurs at the equilibrium position where displacement is zero.
In SHM this is equal to where r is the amplitude of the motion, and will be in this case, the height the mass rises above the equilibrium position from where it started.
You have v and omega. You can find r.
4. So that's the third, and so far the easiest, way

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