A block with mass M attached to a horizontal spring with force constan...
Solution:
Given:
Mass of the block = M
Force constant of the spring = k
Amplitude of the block before putting the lump of putty = A1
Mass of the putty = m
To find:
1. Amplitude and period of the block after putting the lump of putty when it passes through the equilibrium position.
2. Amplitude and period of the block after putting the lump of putty when it is at one end of its path.
1. Lump of putty is dropped on the block at the equilibrium position:
• When the lump of putty is dropped on the block, the system becomes a composite system of M+m mass.
• The new mass of the system becomes M+m.
• The force constant of the spring remains the same.
• Hence, the new angular frequency of the system is given by ω' = √(k/(M+m))
• The new period of the system is given by T' = 2π/ω'
• The total energy of the system remains the same.
• Initially, the energy of the system is given by E = (1/2)kA1^2
• After putting the lump of putty, the amplitude of the system becomes A2.
• The new amplitude can be found using the conservation of energy.
• The total energy of the system after putting the lump of putty is E = (1/2)(M+m)ω'^2A2^2
• Equating the two expressions for energy, we get:
(1/2)kA1^2 = (1/2)(M+m)ω'^2A2^2
• Substituting the values of ω' and solving for A2, we get:
A2 = A1/(√(1+m/M))
2. Lump of putty is dropped on the block when it is at one end of its path:
• When the lump of putty is dropped on the block, the system becomes a composite system of M+m mass.
• The new mass of the system becomes M+m.
• The force constant of the spring remains the same.
• Hence, the new angular frequency of the system is given by ω' = √(k/(M+m))
• The new period of the system is given by T' = 2π/ω'
• The total energy of the system remains the same.
• Initially, the energy of the system is given by E = (1/2)kA1^2
• After putting the lump of putty, the amplitude of the system becomes A3.
• The new amplitude can be found using the conservation of energy.
• The total energy of the system after putting the lump of putty is E = (1/2)(M+m)ω'^2A3^2
• Equating the two expressions for energy, we get:
(1/2)kA1^2 = (1/2)(M+m)ω'^2A3^2
• Substituting the values of ω' and solving for A3, we get:
A3 = A1/(√(1+2(m/M)))
Thus, the new amplitude and period of the block after putting the lump of putty can be calculated using the above equations.
A block with mass M attached to a horizontal spring with force constan...
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