wo springs with negligible masses and force constants k1=200 N/m𝑘1=200 N/m and k2=160 N/m𝑘2=160 N/m are attached to the block of mass m=10 kg𝑚=10 kg as shown in figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time t=0,𝑡=0, sharp impulse of 50 N−s50 N-s is given to the block with a hammer along the spring. Then the amplitude of spring will be......m......m.
Question
wo springs with negligible masses and force constants k1=200 N/m𝑘1=200 N/m and k2=160 N/m𝑘2=160 N/m are attached to the block of mass m=10 kg𝑚=10 kg as shown in figure. Initially the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time t=0,𝑡=0, sharp impulse of 50 N−s50 N-s is given to the block with a hammer along the spring. Then the amplitude of spring will be......m......m.
Solution
The amplitude of the spring can be found by using the principle of conservation of momentum and the principle of conservation of energy.
Step 1: Conservation of Momentum The initial momentum of the block is zero because it is at rest. After the impulse, the momentum of the block is the impulse given, which is 50 N-s. Therefore, the velocity of the block immediately after the impulse is given by:
v = p/m = 50 N-s / 10 kg = 5 m/s
Step 2: Conservation of Energy The total energy in the system is conserved. Immediately after the impulse, all the energy is kinetic energy. At the maximum amplitude, all the energy is potential energy stored in the springs. Therefore, we can equate the kinetic energy and the potential energy to find the amplitude.
Kinetic Energy = 1/2 * m * v^2 = 1/2 * 10 kg * (5 m/s)^2 = 125 J
Potential Energy = 1/2 * k1 * x^2 + 1/2 * k2 * x^2 = 1/2 * (k1 + k2) * x^2
where x is the amplitude. Solving for x gives:
x = sqrt((2 * Kinetic Energy) / (k1 + k2)) = sqrt((2 * 125 J) / (200 N/m + 160 N/m)) = sqrt((250 J) / (360 N/m)) = 0.83 m
Therefore, the amplitude of the spring is 0.83 m.
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