A block oscillating on a spring has an amplitude of 20 cm. What will be the ampl

Redemitz4s 2021-11-22 Answered
A block oscillating on a spring has an amplitude of 20 cm. What will be the amplitude if the maximum kinetic energy is doubled?
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Expert Answer

Fachur
Answered 2021-11-23 Author has 17 answers
Step 1
At equilibrium postion the energy is pure kinetic energy:
Ek=12mvmax2
Step 2
The maximum speed is given by:
vmax=ωA
where ω is angular frequency and A is the amplitude of the spring.
Substitution of the previous relation into the the expression for kinetic energy gives:
Ek=12mA2ω2
Step 3
After doubline the kinetic energy Ek=2Ek, the ratio of the kinetic energies is:
EkEk=12mA2ω212mA2ω2
EkEk=A2A2
thereby:
2=A2A2
A=2A
substituing the given value A=20cm, final result is obtained:
A=28.3cm
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Nancy Johnson
Answered 2021-11-24 Author has 17 answers
Step 1
In the simple harmonic motion, the energy is conserved and the total mechanical energy E has an expression in terms of the force constant k and amplitude A as given by equation (15.21) in the form
E=K+U
12kA2=12mv2+12kx2
The total mechanical energy is
E=12kA2
1) A=2Ek
As shown by equation (1), the amplitude is directly proportional to root square root of E
A=E
As k is constant, we can find the amplitude for two instants when E is doubled by
A1A2=E2E
A1A2=12
A2=2A1
A2=2(20cm)
A2=28.28cm
The amplitude is 28.28 cm when E is doubled.
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