Question

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 4.5 cm has an energy of 26 J. If the block is replaced by one whose mass is twice the mass of the original block and the amplitude of the motion is again 4.5 cm, what is the energy of the system?

Answer 1

Answer:

k=2.157

Spring constant is 2.157 for the amplitude of block spring system

Explanation:

Given

Amplitude of block-spring system is 4.5cm

energy is 26J

EXPRESSION

Total energy of a spring block system on a frictionless horizontal surface

$U=\dfrac{1}{2}kA^2----------------(1)$

Here,

k → spring constant

A → amplitude of the block spring system

Where k is 4.5cm and energy is 26J

Substitute 4.5cm for A

Substitute 26J  for U

Put value in equation 1

$26J=\dfrac{1}{2}k{(4.5cm)^{2}$

k=2.157

Spring constant is 2.157 for the amplitude of block spring system