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Five different experiments are carried out. In each experiment, a block is attac

ID: 2098594 • Letter: F

Question

Five different experiments are carried out. In each experiment, a block is attached to a horizontal spring. The block is pulled back a certain distance and released. The block oscillates back and forth on a frictionless surface. Rank the amplitude of oscillation for each of the following situations. (Rank the smallest amplitude as 1)


A block of mass M is attached to a spring with a spring constant k, pulled back a distance d, and released.

A block of mass (1/2)M is attached to a spring with a spring constant k, pulled back a distance d, and released.

A block of mass M is attached to a spring with a spring constant (1/2)k, pulled back a distance (1/2)d, and released.

A block of mass M is attached to a spring with a spring constant 2k, pulled back a distance d, and released.

A block of mass M is attached to a spring with a spring constant k, pulled back a distance (1/2)d, and released.

Explanation / Answer

The maximum force on the block can be calculated using the Hooke's law                F = kx Here, k is the spring constant and x is the compression or stretched distance i) Spring constant is k , stretched distance is d/2                    F1 = k (d/2)                         =1/2 (kd)           Rank - (1) ii) Spring constant is 2k , stretched distance is 2d                    F2 = 2k (2d)                         = 4 (kd)             Rank - (3) iii) Spring constant is 1/2k , stretched distance is d                    F3 = 1/2k (d)                         = 1/2 (kd)         Rank - (1) iv)Spring constant is k , stretched distance is d                    F4 = k (d)                         = kd                   Rank - (2) v)Spring constant is k , stretched distance is d                    F5 = k (d)                         = kd                   Rank - (2) F2 > (F4 =  F5) > ( F1 = F3 )                       F3 = 1/2k (d)                         = 1/2 (kd)         Rank - (1) iv)Spring constant is k , stretched distance is d                    F4 = k (d)                         = kd                   Rank - (2) v)Spring constant is k , stretched distance is d                    F5 = k (d)                         = kd                   Rank - (2) F2 > (F4 =  F5) > ( F1 = F3 )                       F4 = k (d)                         = kd                   Rank - (2) v)Spring constant is k , stretched distance is d                    F5 = k (d)                         = kd                   Rank - (2) F2 > (F4 =  F5) > ( F1 = F3 )                       F5 = k (d)                         = kd                   Rank - (2) F2 > (F4 =  F5) > ( F1 = F3 )