Question

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4.) A harmonic oscillator starts with an amplitude of 20.0 cm. After 10.0 s, the amplitude decreases to 15.0 cm. If the linear damping coefficient is 2.00 Ns/m, how much mass is oscillating?

Answer 1

As we know that the equation of motion of damped oscillation is given by the second order
differential equation is

( d ² x / dt ² ) + 2 β ( d x / dt ) + ω₀ ² x = 0

where -
β = b / ( 2m ) [ b = damping constant, m = mass]

ω₀ = √ [ k / m ]
k = spring constant

Now assume small damping => underdamped oscillation -

The corresponding solution of the differential equation is given by -

x ( t ) = A₀ exp [ - β t ] cos ( ω₁ t – δ ) = A cos ( ω₁ t – δ )

Where -

A = A₀ exp [ - β t ]

ω₁ = √ [ ω₀ ² - β ² ]

δ = phase constant
Since the exponential function gives numbers only, the unit of β must be 1 / s

β = b / ( 2 m )

=> b = 2 m β

(Therefore, unit of the damping constant (coefficient) b should be ( kg / s ), not N/m = unit for spring constant)

Given that,

b = 2 kg / s, A₀ = 20 cm, A = 15 cm and t = 10 s

So,

A / A₀ = 15 cm / 20 cm = 0.75 = exp [ - β ( 10 s ) ]

Take the logarithmic function -

ln ( 0.75 ) = - 0.287682 = - β ( 10 s )

=> β = 0.287682 / ( 10 s ) = 0.0287682 s ⁻ ¹

Now, β = b / ( 2 m )

=> m = b / [ 2 β ] = [ 2 kg / s ] / [ 2 ( 0.0287682 s ⁻ ¹ ) ] = 34.8 kg

Hence the oscillating mass = 34.8 kg.