Two students on ice skates stand one behind the other. Student 2 pushes student 1 in the back; both students move away from each
1 answer:
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a)
The equation which gives the position of a simple harmonic oscillator is:
where
A is the amplitude
is the angular frequency, with k being the spring constant and m the mass
t is the time
Let's start by calculating the angular frequency:
The amplitude, A, can be found from the maximum velocity of the spring:
So, the equation of motion is
b) t=0.10 s, t=0.52 s
The potential energy is given by:
While the kinetic energy is given by:
The velocity as a function of time t is:
The problem asks as the time t at which U=3K, so we have:
However, , so we have
with two solutions:
c) 3 seconds.
When x=0, the equation of motion is:
so, t=0.
When x=1.00 m, the equation of motion is:
So, the time needed is 3 seconds.
d) 0.097 m
The period of the oscillator in this problem is:
The period of a pendulum is:
where L is the length of the pendulum. By using T=0.628 s, we find
Answer:
Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?
Explanation:
So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s
so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull