Find Amplitude and Frequency

  • Thread starter dsptl
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  • #1
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* A 200g block attached to a horizontal spring is oscillating with an amplitude of 3 cm and a frequency of 2 Hz. Just as it passes through the equilibrium point a shalock so trp blow exerts a force impulse on the that the speed of the block is increased by .5m/s. What are the new a) frequency and b) amplitude?

My Attempt:

total energy of the mass moving with amplitude A and oscillating with frequency f will be
E = mA^2ω^2 / 2

where ω = 2πf
f = frequancy of oscillation
A = amplitude of oscillation .
at the mean postion the total eneregy will be in kinetic energy form
E = mv^2 / 2
v =velocity of the mass at equalibrium
if 0.5m/sec is added to velocity

Now the new total eneryg
E' = m*(v + 0.5)^2 / 2 .
m = mass of the block
E' = mA'^2 ω^2 / 2
the frequancy remains constanat
and the amplidue to of the new oscillation changes
ω = 2E' / mA'^2


but at the end I am left with 2 unknown and i am not to solve this


please help its urgent
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi dsptl,

* A 200g block attached to a horizontal spring is oscillating with an amplitude of 3 cm and a frequency of 2 Hz. Just as it passes through the equilibrium point a shalock so trp blow exerts a force impulse on the that the speed of the block is increased by .5m/s. What are the new a) frequency and b) amplitude?

My Attempt:

total energy of the mass moving with amplitude A and oscillating with frequency f will be
E = mA^2ω^2 / 2

where ω = 2πf
f = frequancy of oscillation
A = amplitude of oscillation .
at the mean postion the total eneregy will be in kinetic energy form
E = mv^2 / 2
v =velocity of the mass at equalibrium
That's right; so what is the speed of the mass at the equilibrium point before the force impulse add the 0.5m/s? Then you will be able to find the value of the energy after the force impulse. What do you get?
 
  • #3
64
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total energy of the mass moving with amplitude A and oscillating with frequency f will be
E = mA^2ω^2 / 2 = (.2)(.03)^2(4pi)^2 / 2 = .0142J

where ω = 2πf
f = frequancy of oscillation
A = amplitude of oscillation .

at the mean postion the total eneregy will be in kinetic energy form
E = mv^2 / 2
.0142 = .2 v^2 /2

v = .377m/s

v =velocity of the mass at equalibrium
if 0.5m/sec is added to velocity = .377m/s + .5m/s = .877m/s


Is this right till this point? after this point I have no idea what to do....

Now the new total eneryg
E' = m*(v + 0.5)^2 / 2 .
m = mass of the block
E' = mA'^2 ω^2 / 2
the frequancy remains constanat
and the amplidue to of the new oscillation changes



.
 
  • #4
alphysicist
Homework Helper
2,238
1
total energy of the mass moving with amplitude A and oscillating with frequency f will be
E = mA^2ω^2 / 2 = (.2)(.03)^2(4pi)^2 / 2 = .0142J

where ω = 2πf
f = frequancy of oscillation
A = amplitude of oscillation .

at the mean postion the total eneregy will be in kinetic energy form
E = mv^2 / 2
.0142 = .2 v^2 /2

v = .377m/s

v =velocity of the mass at equalibrium
if 0.5m/sec is added to velocity = .377m/s + .5m/s = .877m/s


Is this right till this point? after this point I have no idea what to do....

Now the new total eneryg
E' = m*(v + 0.5)^2 / 2 .
m = mass of the block
E' = mA'^2 ω^2 / 2
the frequancy remains constanat
and the amplidue to of the new oscillation changes
Perhaps I'm not understanding what you're asking, but you now can solve for A', since you have everything else.
 

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