Question

A mas-spring system is vibrating on a frictionless, horizontal surface with an amplitude of 6.0 cm. The system has a total mechanical energy of 12 J. If the block is replaced by one whose mass is three times the mass of the original block (m2 = 3m1) and the amplitude of the motion is again 6.0 cm, what is the new maximum velocity of the system?

Answer 1

Answer:

given,

mass of the block = m₁

mass of the another block = 3 m₁

initial Amplitude, A = 6 cm

final amplitude = 6 cm

total mechanical energy = 12 J

total energy of the block spring

   $E = \dfrac{1}{2}kA^2$

A is the amplitude and k is spring constant

initial energy is equal to 12 J

from the above expression we can say that

Energy of the given system depends up on the magnitude of spring constant and the amplitude.

so, energy of both the system will be same.

we know,

$E = \dfrac{1}{2}mv^2$

$12= \dfrac{1}{2}\times 3 m_1 v^2$

  $v^2 = \dfrac{8}{m_1}$

  $v = \sqrt{ \dfrac{8}{m_1}}$