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iren [92.7K]
2 years ago
12

A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app

lied a damping force (in Newtons) of the form F = –0.1v, with v the velocity in m/s. The spring is stretched a distance xm and released. After four complete oscillations, what fraction of the mechanical energy is retained by the system?

Physics
1 answer:
Cerrena [4.2K]2 years ago
6 0

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

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Answer:

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7  m/s

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The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7  m/s.

8 0
2 years ago
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

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The quantity of heat transferred per unit mass is given by;

Q = cΔT

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Q(A) = caΔT

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ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

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The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

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T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

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2 years ago
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