a 1kg block is executing simple harmonic motion of amplitude 0.1m on a smooth horizontal surface under the restoring force of a spring of spring constant 100N/M.A block of mass 3kg is gently placed on it at the instant it passes through the mean position .Assuming that the two blocks move togther,find the frequency and the amplitude of the motion?
Before the 3 kg weight was added, the period was
P = 2 pi sqrt(m/k) = 0.628 s.
When you add 3 km, the total mass is quadrupled and the period is increased by a facgtor of sqrt4 = 2. The new period is
P = 1.257 s.
The new frequency is 1/P = 0.796 Hz
With the mass added at the equilibrium position, the velocity instantly drops by a factor of 4 to preserve linerar momentum. The total kinetic energy at that point is then reduced by a factor (4/1)*(1/4)^2 = 1/4
With the new reduced total energy, the pair of blocks can only have an amplitude sqrt (1/4) = 1/2 as large as before.
To find the frequency and the amplitude of the motion, we can apply the principles of simple harmonic motion.
Let's start by considering the system before the 3kg block is placed on the 1kg block. The restoring force on the 1kg block is provided by the spring, and we can use the formula for the angular frequency (ω) of the motion:
ω = √(k / m)
where k is the spring constant and m is the mass of the 1kg block.
Substituting the given values, we have:
ω = √(100 N/m / 1 kg)
= √(100 rad^2/s^2)
= 10 rad/s
The angular frequency (ω) is related to the frequency (f) by the equation:
ω = 2πf
Substituting ω = 10 rad/s, we can solve for f:
10 rad/s = 2πf
f = 10 rad/s / 2π
f ≈ 1.59 Hz
Now, let's consider the system after placing the 3kg block on the 1kg block. The mass of the combined system is now 4kg (1kg + 3kg), but the spring constant remains the same.
To find the new angular frequency (ω') and amplitude (A') of the motion, we can use the principle of conservation of momentum at the mean position, assuming the blocks move together.
The momentum before the 3kg block is placed is given by the product of the mass (1kg) and velocity at the mean position (v):
momentum before = 1kg * v
The momentum after the blocks move together is given by the product of the combined mass (4kg) and the new velocity at the mean position (v'):
momentum after = 4kg * v'
Since momentum is conserved, we can equate the two expressions:
1kg * v = 4kg * v'
Solving for v', we have:
v' = v / 4
The amplitude of the new motion is half of the displacement of the 1kg block, which is 0.1m. So, the amplitude of the motion for the combined system is:
A' = 0.1m / 2
A' = 0.05m
Now, we can find the new angular frequency (ω') using the formula:
ω' = √(k / m')
where k is the spring constant and m' is the combined mass of the system.
Substituting the values, we have:
ω' = √(100 N/m / 4 kg)
= √(25 rad^2/s^2)
= 5 rad/s
The new frequency (f') is given by:
ω' = 2πf'
5 rad/s = 2πf'
f' = 5 rad/s / 2π
f' ≈ 0.796 Hz
Therefore, the frequency of the motion when the 3kg block is placed on the 1kg block is approximately 0.796 Hz and the amplitude is 0.05m.