A loudspeaker diaphragm is vibrating in simple harmonic motion with a frequency of 430 Hz and a maximum displacement of 0.90 mm.
what is the max speed?
what is the max acceleration?
T = period = 1/430
for example if 0 at t = 0
y = .009 sin [2pi t/(1/430)] = .0009 sin( 860 pi t)
velocity = .0009 (860 pi) cos (860 pi t)
acceleration = -.0009 (860 pi)^2 sin(860 pi t)
the maximum of sine is 1 and the maximum of cosine is 1
To find the maximum speed and maximum acceleration of the vibrating loudspeaker diaphragm, we can use the equations of motion for an object undergoing simple harmonic motion.
The maximum velocity in simple harmonic motion occurs when the displacement is maximum. In this case, the maximum displacement is given as 0.90 mm. The relationship between maximum displacement and maximum velocity is given by:
v_max = ω * A
Where:
v_max is the maximum velocity,
ω (omega) is the angular frequency, and
A is the amplitude (maximum displacement).
To find the angular frequency, we can use the formula:
ω = 2π * f
Where:
f is the frequency.
Plugging in the values:
f = 430 Hz
A = 0.90 mm (which needs to be converted to meters)
First, let's convert the amplitude to meters:
A = 0.90 mm = 0.90 × 10^(-3) m
Now, let's find the angular frequency:
ω = 2π * f = 2π * 430 Hz ≈ 2703.57 rad/s
Finally, we can find the maximum velocity:
v_max = ω * A = 2703.57 rad/s * 0.90 × 10^(-3) m ≈ 2.43 m/s
Therefore, the maximum speed of the loudspeaker diaphragm is approximately 2.43 m/s.
To find the maximum acceleration, we can use the equation:
a_max = ω^2 * A
Plugging in the values we already know:
ω = 2703.57 rad/s
A = 0.90 × 10^(-3) m
Let's calculate the maximum acceleration:
a_max = ω^2 * A = (2703.57 rad/s)^2 * 0.90 × 10^(-3) m ≈ 6.56 m/s²
Therefore, the maximum acceleration of the loudspeaker diaphragm is approximately 6.56 m/s².