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riadik2000 [5.3K]
3 years ago
7

The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at t

he moments of time: 0, T, 2T, 3T and 4T.​

Physics
2 answers:
zhannawk [14.2K]3 years ago
8 0

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

Salsk061 [2.6K]3 years ago
6 0

Answer:

Explanation:

ω = π/2 = 2π/T

T = 2π/(π/2) = 4 s

velocity is the derivative of position

v(t) = (-1/80)(e^(-t/4)[sin(πt/2) - 2πcos(πt/2)]

0T = v(0) = (-1/80)(e^(-0/4)[sin(π0/2) - 2πcos(π0/2)] = 0.078540

 T = v(4) = (-1/80)(e^(-4/4)[sin(π4/2) - 2πcos(π4/2)] = 0.028893

2T = v(8) = (-1/80)(e^(-8/4)[sin(π8/2) - 2πcos(π8/2)] = 0.010629

3T = v(12) = (-1/80)(e^(-12/4)[sin(π12/2) - 2πcos(π12/2)] = 0.003910

4T = v(16) = (-1/80)(e^(-16/4)[sin(π16/2) - 2πcos(π16/2)] = 0.001439

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Answer:1.04 N

Explanation:

Given

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This gravitational Force has components along and Perpendicular to Platter

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W_p=5\times \cos 12=4.89 N

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3 years ago
While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

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