Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.
What is the Acceleration of a Body Directly Proportional to?
Let the earth be considered as a sphere of uniform density (same density at every part of the earth). Let AB be a diameter [Fig.] of the earth. A frictionless tunnel is imagined along AB. (Such a tunnel has no physical existence.) To show the characteristic features of the acceleration due to gravity below the surface of the earth, such a tunnel is imagined.
It can be shown that on releasing a body through the tunnel, the body will execute a simple harmonic motion.
Let a body of mass m be dropped in the tunnel AB. After some time, it reaches the point C, at a distance x from the centre of the earth. Taking O as the centre and OC = x as the radius, a sphere is drawn. Here only the gravitational force due to the sphere of radius x is effective and the spherical shell of thickness AC(= R – x) applies no force on the body.
Hence, the force of attraction on the body at C,
F \(=\frac{G \times \text { mass of sphere of radius } x \times m}{x^2}\)
= G ᐧ \(\frac{\frac{4}{3} \pi x^3 \rho m}{x^2}\) (ρ = average density of the earth)
= \(\frac{4}{3}\)πGρmx
∴ Acceleration produced a = \(\frac{F}{m}\) = \(\frac{4}{3}\)πGρx ……….. (1)
Hence, a ∝ x [as \(\frac{4}{3}\)πGρ = constant]
Therefore,
- the acceleration of the body is directly proportional to its distance from the centre of the earth and
- it is always directed towards the centre of the earth.
Since, the gravitational force of attraction is always directed towards the centre of the earth, but the distance is always measured away from the centre of the earth, we may write, a ∝ -x.
Whenever the acceleration of a body with respect to a fixed point fulfills conditions (i) and (ii) simultaneously the body executes simple harmonic motion.
Time period of this motion,
Hence, the time period does not depend on the mass of the body.
Taking the radius of the earth R = 6400 km and the acceleration due to gravity on the surface of the earth g = 9.8 m ᐧ s-2,
T = 2π\(\sqrt{\frac{6400 \times 10^3}{9.8}}\) = 1 h 24 min 38s
The body will move from A to B in exactly half of the above calculated time. Thus, any body, from a certain point on the earth’s surface, will reach exactly the opposite end point in about 42 min.
If the tunnel were not through the centre, but along a straight line joining any two points on the earth’s surface, the body would still have executed simple harmonic motion of the same period and would have taken 42 min to reach one end from the other.
Numerical Examples
Example 1.
The normal length of a steel spring is 8 cm. Keeping one end of the spring fixed at a point, if a weight is attached to the other end, its length becomes 14 cm. The weight is pulled down slightly and then released. Find the time period of oscillation of the spring?
SoIuton:
Increase in length of the spring for the mass m is, l = 14 – 8 = 6 cm
So, force constant k (force required for a unit increase in length) = \(\frac{m g}{6}\) dyn ᐧ cm-1.
Time period, T = 2π\(\sqrt{\frac{m}{k}}\) = 2π\(\sqrt{\frac{m}{\frac{m g}{6}}}\)
= 2π\(\sqrt{\frac{6}{g}}\) = 2 × 3.14 × \(\sqrt{\frac{6}{980}}\) = 0.49 s.
Example 2.
Two bodies of mass m1 and m2 are suspended from a weightless spring. The force constant of the spring is k. When the bodies are in equilibrium position, the body of mass in1 is taken away from the system such that the equilibrium condition of the system is not disturbed at that very moment. Determine the angular frequency and the amplitude of motion for the body of mass m2.
Solution:
If increase in length of the spring due to the two masses m1 and m2 is l, then
k = \(\frac{\left(m_1+m_2\right) g}{l}\) or, l = \(\frac{\left(m_1+m_2\right) g}{k}\)
Similarly for the mass m2, increase in length, l2 = \(\frac{m_2 g}{k}\). This l2 is the increase in length for the final equilibrium position. So with the mass m1, displacement from the equilibrium position = amplitude of motion
= l – l2 = \(\frac{\left(m_1+m_2\right) g}{k}\) – \(\frac{m_2 g}{k}\) = \(\frac{m_1 g}{k}\)
Since only the mass m2 vibrates, the angular frequency ω is given by,
or, ω2 = \(\frac{k}{m_2}\) or, ω = \(\sqrt{\frac{k}{m_2}}\)
Example 3.
A spring is elongated by 2 cm due to a 80 g mass attached to it. Another body of mass 600 g is attached to the end of the spring and it is displaced by 8 cm from its equilibrium position. Calculate the energy of the system in this position. Considering the principle of conservation of energy, determine the velocity of the body when it is at a distance of 4 cm.
Solution:
Force constant of the spring,
k = \(\frac{80 \times 980}{2}\) = 40 × 980 dyn ᐧ cm-1
Mass, m = 600 g; amplitude, A = 8 cm.
Total energy, E = maximum potential energy potential energy at the ends of the path of motion = \(\frac{1}{2}\)kA2 = \(\frac{1}{2}\) × 40 × 980 × (8)2 = 1254400 erg = 0.12544 J.
Even for x = 4 cm, the total energy remains unchanged. If v is the velocity at this position, then
Example 4.
A particle is executing SHM. If time is measured from when it is at one end of its path of motion, calculate the ratio of its kinetic energy to the potential energy at t = \(\frac{T}{12}\). Here T is the time period of the the initial phase is zero.
Solution:
If time is measured from when the particle is at one end of the path of motion, then the equation of SHM is x = Acosωt.
Example 5.
When a man of mass 60 kg sits inside a car, the centre of gravity of the car descends by 0.3 cm. If the mass of the car is 1000 kg, calculate the frequency of oscillation of the empty car.
Solution:
Frequency of oscillation of the car, n = \(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\).
The force constant of the spring of the car,
Example 6.
A wooden block of cross-sectional area 10 cm2 is floating vertically on water. The volume of the immersed portion of the block is 200 cm3. The block is depressed slightly Inside water and then released. Calculate the time period of vibration of the block.
Solution:
Volume of displaced water = 200 cm3
∴ Mass of displaced water = 200 g, and mass of the wooden block = 200g
Let the block be depressed inside water through x cm and then released.
∴ Upward restoring force on the block,
F = 10x × 1 × g = 10xg
∴ Acceleration of the block,
a = \(\frac{F}{m}\) = \(\frac{10 x g}{200}\) = \(\frac{x g}{20}\)
∴ Time period of vibration of the block,
T = 2π\(\sqrt{\frac{x}{a}}\) = 2π\(\sqrt{\frac{20}{g}}\) = 2π\(\sqrt{\frac{20}{980}}\) = 0.897 s.
Example 7.
A small coin is kept on a horizontal platform. The platform is oscillating vertically with a time period of 0.5 s. What should be the maximum amplitude of vibration so that the coin always remains in contact with the platform?
Solution:
The coin will always remain in contact with the platform if the downward acceleration of the platform does not exceed the acceleration of the coin due to gravity.
If A is the maximum amplitude of vibration, then
ω2A = g
or, A = \(\frac{g}{\omega^2}\) = \(\frac{g}{\left(\frac{2 \pi}{T}\right)^2}\) = \(\frac{g T^2}{4 \pi^2}\)
= \(\frac{9.8 \times(0.5)^2}{4 \times(3.14)^2}\) = 0.06205 m ≈ 0.06 m.
Example 8.
Two identical bodies, each of mass m, are connected by a spring having spring constant k and they are placed on a frictionless floor. The spring is compressed a little and then released. What will be the frequency of oscillation of the system? [WBJEE 2000]
Solution:
If the maximum compression of the spring from its position of equilibrium is A, then restoring force = -kA. In this condition, the whole energy of the spring is its potential energy = \(\frac{1}{2}\)kA2.
Again, during oscillation, when the two bodies just cross thee position of equilibrium, the potential energy becomes zero and the total energy is then equal to the kinetic energy of the two bodies. At this stage, velocity of each body maximum
velocity = ω\(\frac{A}{2}\), where ω = angular frequency and \(\frac{A}{2}\) = amplitude of vibration of each body.
∴ Kinetic energy of the two bodies
= \(\frac{1}{2} m \omega^2\left(\frac{A}{2}\right)^2\) + \(\frac{1}{2} m \omega^2\left(\frac{A}{2}\right)^2\) = \(\frac{1}{4} m \omega^2 A^2\)
According to the principle of conservation of energy,
\(\frac{1}{2} k A^2\) = \(\frac{1}{4} m \omega^2 A^2\) or, ω = \(\sqrt{\frac{2 k}{m}}\)
∴ Frequency of oscillation 01 the system = \(\frac{\omega}{2 \pi}\) = \(\frac{1}{2 \pi} \sqrt{\frac{2 k}{m}}\).
Example 9.
The time period of a body of mass M executing SHM, connected to a spring, is 2 s. If the mass of the body is increased by 2 kg, its time period increases by 1 s. Considering that Hooke’s law is obeyed, calculate the Initial mass M.
Solution:
The time period of SHM executed by the body connected to the spring, T = 2π\(\sqrt{\frac{M}{k}}\); k = force constant of the spring
So, in the first case,
2 = 2π\(\sqrt{\frac{M}{k}}\) …. (1)
and in the second case,
2 + 1 = 3 = 2π\(\sqrt{\frac{M+2}{k}}\) ….. (2)
Dividing (2) by (1) we get,
\(\frac{3}{2}\) = \(\sqrt{\frac{M+2}{M}}\) or, \(\frac{9}{4}\) = \(\frac{M+2}{M}\)
or, 9M = 4M + 8 or, M = 1.6 kg.
Example 10.
Time period of a spring of negligible mass, with a mass M hanging from it, is T. The time period changes to \(\frac{5 T}{3}\) on attaching an additional mass m to it. Find out the value of \(\frac{m}{M}\). [AIEEE ‘03]
Solution: