An air chamber of volume V has a neck of cross-sectional area a into which a light ball of mass m just fits and can move up and down without friction. The diameter of the ball is equal to that of the neck of the chamber. The ball is pressed down a little and released. If the bulk modulus of air is B, the time period of the oscillation of the ball is
Answers
Answer:
Volume of the air chamber = V
Area of cross-section of the neck = a
Mass of the ball = m
The pressure inside the chamber is equal to the atmospheric pressure.
Let the ball be depressed by x units.
As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.
Decrease in the volume of the air chamber, ΔV = ax
Volumetric strain = Change in volume/original volume
⇒ ΔV/V = ax/V
Bulk modulus of air, B = Stress/Strain = fraction numerator straight rho over denominator begin display style bevelled ax over straight V end style end fraction
In this case, stress is the increase in pressure.
The negative sign indicates that pressure increases with decrease in volume.
p = -Bax/V
The restoring force acting on the ball, F = p × a
equals space minus Bax over straight V space cross times space a space
equals space minus fraction numerator B a x squared over denominator V end fraction ...(i)
In simple harmonic motion, the equation for restoring force is,
F = -kx ...(ii)
where, k is the spring constant.
Comparing equations (i) and (ii), we get:
k = Ba2/V
Time Period,.