254
6 Oscillations
6.60 A damped oscillator loses 3% of its energy in each cycle. (a) How many cycles
elapse before half its original energy is dissipated? (b) What is the Q factor?
6.61 A damped oscillator has frequency which is 9/10 of its natural frequency. By
what factor is its amplitude decreased in each cycle?
6.62 Show that for small damping ω ≈ (1 − r 2 /8mk)ω 0 where ω 0 is the natural
angular frequency, ω the damped angular frequency, r the resistance constant,
k the spring constant and m the particle mass.
6.63 Show that the time elapsed between successive maximum displacements of a
damped harmonic oscillator is constant and equal to 4π m/
√
4km − r 2 , where
m is the mass of the vibrating body, k is the spring constant, 2b = r/m, r
being the resistance constant.
6.64 A dead weight attached to a light spring extends it by 9.8 cm. It is then slightly
pulled down and released. Assuming that the logarithmic decrement is equal
to 3.1, find the period of oscillation.
6.65 The position of a particle moving along x-axis is determined by the equation
d 2 x/dt 2 + 2dx/dt + 8x = 16 cos 2t.
(a) What is the natural frequency of the vibrator?
(b) What is the frequency of the driving force?
6.66 Show that the time t 1/2 for the energy to decrease to half its initial value is
related to the time constant by t 1/2 = t c ln 2.
6.67 The amplitude of a swing drops by a factor 1/e in 8 periods when no energy
is pumped into the swing. Find the Q factor.
6.3 Solutions
6.3.1 Simple Harmonic Motion (SHM)
6.1 x = A sin ωt
(SHM)
ω =
2π
T
=
2π
2π
= 1 rad/s
8
√
2 = A sin
1 · π
4
A = 16 cm = 0.16 m
E =
1
2
m A
2
ω
2
∴ m =
2E
A 2 ω 2 =
2 × 0.256
(0.16) 2 × 1 2 = 20.0 kg
6 Oscillations
6.60 A damped oscillator loses 3% of its energy in each cycle. (a) How many cycles
elapse before half its original energy is dissipated? (b) What is the Q factor?
6.61 A damped oscillator has frequency which is 9/10 of its natural frequency. By
what factor is its amplitude decreased in each cycle?
6.62 Show that for small damping ω ≈ (1 − r 2 /8mk)ω 0 where ω 0 is the natural
angular frequency, ω the damped angular frequency, r the resistance constant,
k the spring constant and m the particle mass.
6.63 Show that the time elapsed between successive maximum displacements of a
damped harmonic oscillator is constant and equal to 4π m/
√
4km − r 2 , where
m is the mass of the vibrating body, k is the spring constant, 2b = r/m, r
being the resistance constant.
6.64 A dead weight attached to a light spring extends it by 9.8 cm. It is then slightly
pulled down and released. Assuming that the logarithmic decrement is equal
to 3.1, find the period of oscillation.
6.65 The position of a particle moving along x-axis is determined by the equation
d 2 x/dt 2 + 2dx/dt + 8x = 16 cos 2t.
(a) What is the natural frequency of the vibrator?
(b) What is the frequency of the driving force?
6.66 Show that the time t 1/2 for the energy to decrease to half its initial value is
related to the time constant by t 1/2 = t c ln 2.
6.67 The amplitude of a swing drops by a factor 1/e in 8 periods when no energy
is pumped into the swing. Find the Q factor.
6.3 Solutions
6.3.1 Simple Harmonic Motion (SHM)
6.1 x = A sin ωt
(SHM)
ω =
2π
T
=
2π
2π
= 1 rad/s
8
√
2 = A sin
1 · π
4
A = 16 cm = 0.16 m
E =
1
2
m A
2
ω
2
∴ m =
2E
A 2 ω 2 =
2 × 0.256
(0.16) 2 × 1 2 = 20.0 kg