Question

If the object-spring system is described by *x* = (0.340
m) cos (1.55*t*), find the following.

(a) the amplitude, the angular frequency, the frequency, and the period

A = |
m |

ω = |
rad/s |

f = |
Hz |

T = |
s |

(b) the maximum magnitudes of the velocity and the acceleration

v_{max} = |
m/s |

a_{max} = |
m/s^{2} |

(c) the position, velocity, and acceleration when *t* =
0.250 s

x = |
m |

v = |
m/s |

a = |
m/s^{2} |

Answer #1

Displacement of a spring in simple harmonic motion is given
by,

x(t) = Acos(t + )

where x is the position

A is the amplitude

is the angular
frequency

is
the phase

For this problem,

x = (0.340 m) cos (1.55t)

Just find the corresponding numbers.

A = 0.340 m

= 1.55
rad/s

The frequency is given by

f = /2

f = (1.55 rad/s) / 2

f = 0.246 Hz

The period is just the inverse of the frequency.

T = 1/f

T = 1/(0.246 Hz)

T = 4.05 s

B)Next we need to find the velocity and acceleration,

v(t) = -Acos(t +
)

and

a(t) = A^{2}cos(t +)

Then maximum velocity is A, and the maximum
acceleration is A^{2}.

The maximum velocity is v_{max}

v_{max} = A = (0.340 m)(1.55
rad/s) = 0.527 m/s

a_{max}= A^{2}
=(0.340 m)(1.55 rad/s)^{2} =0.82 m/s²

c) Plug in the value for t, let us assume the phase =0.

x(t) = Acos(t + )

x(0.250 s) = (0.340 m)cos((1.55 rad/s)(0.250 s) + 0 rad)

x(0.250 s) = 0.339 m

v(t) = -Acos(t +
)

v(0.250 s) = -(0.340 m)(1.55 rad/s)cos((1.55 rad/s)(0.250 s) + 0
rad)

v(0.250 s) = -0.526 m/s

a(t) = A^{2}cos(t + )

a(0.250 s) = (0.340 m)(1.55 rad/s)^{2}cos((1.55
rad/s)(0.250 s) + 0 rad)

a(0.250 s) = 0.817 m/s^{2}

If the object-spring system is described by x = (0.310
m) cos (1.55t), find the following.
(a) the amplitude, the angular frequency, the frequency, and the
period
A =
m
ω =
The angular frequency is ω in
Acosωt. rad/s
f =
Hz
T =
s
(b) the maximum magnitudes of the velocity and the
acceleration
vmax =
m/s
amax =
m/s2
(c) the position, velocity, and acceleration when t =
0.250 s
x =
m
v =
m/s
a...

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(a) the amplitude, the angular frequency, the frequency, and the
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A=.315 m
w=1.6 rad/s
f=.25 HZ
T=3.9s
(b) the maximum magnitudes of the velocity and the
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vmax=.504 m/s
amax=.8064 m/s^2
(c) the position, velocity, and acceleration when t =
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x= ?
v=?
a=?

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displaced 0.1m to the right of its equilibrium position. Its
initial velocity is 0.4 m s , toward the right.
a) Calculate the period T of the motion. b) Calculate the
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A spring of negligible mass stretches 3.00 cm from its relaxed
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on a frictionless horizontal surface and is attached to the free
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(Assume that the direction of the initial displacement is positive.
Use the exact values you enter to make later calculations.)
(a)...

A 6.5-kg mass is attached to an ideal 750-N/m spring. If the
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The frequency, f =
The angular frequency, ω =
The period, T =
If the total mechanical energy of the system is 72 J, what are
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The amplitude, A =
The maximum speed, vmax =
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The position of a 0.30 kg object attached to a spring is
described by x = (0.25 m) cos(0.4 π t).
(a) What is the amplitude of the motion?
(b) Calculate the spring constant.
(c) Calculate the position of the object at t = 0.30 s.
(d) Calculate the velocity of the object at t = 0.30 s.

The position of a 0.30 - kg object attached to a spring is
described by x = (0.25 m) cos ( 0.4 pie t). Find: a) The amplitude
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The position of a 0.30 - kg object attached to a spring is
described by
x = (0.25 m) cos ( 0.4 π t ).
Find:
a) The amplitude of the motion;
b) Frequency and Period of the motion;
c) Position of the object at t = .30 sec.
d) Speed of the object at t = .30 sec.
e) Acceleration of the object at t = .30 sec

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(a) oscillator's maximum velocity (in m/s)
(b) oscillator's maximum acceleration (m/s2)
(c) oscillator's position (in m) when t = 1.25 s
(d) oscillator's velocity (in m/s) when t = 1.25 s
(e) oscillator's acceleration (in m/s2) when
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The position of a 0.30 - kg object attached to a spring is
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