An oscillator consists of a block attached to a spring (k = 393 N/m). At some time t, the pos...?

An oscillator consists of a block attached to a spring (k = 393 N/m). At some time t, the position (measured from the system's equilibrium location), velocity, and acceleration of the block are x = 0.121 m, v = -16.3 m/s, and a = -106 m/s2. Calculate (a) the frequency (in Hz) of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.

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  • 7 years ago
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    k = spring constant = 393 N/m

    and at a certain (unknown) time t

    x = displacement = 0.121 m

    v = velocity = -16.3 m/s

    a = acceleration = -106 m/s²

    The angular frequency of the motion can be found by:

    a = -xω²

    so

    ω = √( -a / x )

    ω = √( -(-106 m/s²) / (0.121 m) )

    ω = 29.6 rad/s

    The cyclic frequency can be found by:

    ω = 2πf

    so

    f = ω / 2π

    f = (29.6 rad/s) / (2π rad)

    f = 4.71 Hz < - - - - - - - - - - - - - - - - - - - - - answer (a)

    The mass of the block can be found by:

    ω = √( k / m )

    so

    m = k / ω²

    m = (393 N/m) / (29.6 rad/s)²

    m = 0.449 kg < - - - - - - - - - - - - - - - - - - - - - answer (b)

    The kinetic energy of the mass at time t is:

    KE = m × v² / 2

    KE = (0.449 kg) × (-16.3 m/s)² / 2

    KE = 59.6 J

    The potential energy in the spring at time t is:

    PE = k × x² / 2

    PE = (393 N/m) × (0.121m/s)² / 2

    PE = 2.88 J

    The total energy in the systemm is:

    TE = KE + PE

    TE = (59.6 J) + (2.88 J)

    TE = 62.5 J

    The amplitude can be found by:

    TE = k × A² / 2

    so

    A = √( 2 × TE / k )

    A = √( 2 × (62.5 J) / (393 N/m) )

    A = 0.564 m < - - - - - - - - - - - - - - - - - - - - - answer (c)

  • 4 years ago

    thank u

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