When a mass m is hung from a spring of negligible mass with a force constant k, it extends by an amount y. The mass is gently drawn down and released. The system begins to execute simple harmonic motion of amplitude A and angular frequency ω. The total energy of the mass - spring system will beA) 1 2 m A 2 ω 2 B) 1 2 m A 2 ω 2 + 1 2 k y 2 C) 1 2 k y 2 D) 1 2 m A 2 ω 2 − 1 2 k y 2

Question

When a mass m is hung from a spring of negligible mass with a force constant k, it extends by an amount y. The mass is gently drawn down and released. The system begins to execute simple harmonic motion of amplitude A and angular frequency ω. The total energy of the mass - spring system will beA)       1    2    m      A    2        ω    2  B)       1    2    m      A    2        ω    2    +      1    2    k      y    2  C)       1    2    k      y    2  D)       1    2    m      A    2        ω    2    −      1    2    k      y    2

Alleryexerimivn3 · Accepted Answer

The correct answer is B)       1    2    m      A    2        ω    2    +      1    2    k      y    2  Let L be the relaxed length of the spring and y the extension produced in it due to force mg so that ky = mg (i)The displacement of the mass during oscillation is given byx=Asin(ωt+ϕ)At the instant when the displacement is x​KE of mass   =      1    2    m      V    2    =      1    2    m  (            d      x              d      t            )    2    =      1    2    m      A    2        ω    2        cos    2    ⁡  (  ω  t  +  ϕ  ) (iii)PE of spring   =      1    2    m      V    2    =      1    2    m  (            d      x              d      t            )    2    =      1    2    m      A    2        ω    2        cos    2    ⁡  (  ω  t  +  ϕ  )Using (i) and (ii) and   ω  =            k      m      , we havePE of spring   =      1    2    k      y    2    +  m  g  x  +      1    2    m      ω    2        A    2        sin    2    ⁡  (      ω    t  +      ϕ    ) (iv)taking gravitational PE at the mean position to be zero.Gravitational PE at x = -mg x (v)Adding (iii), (iv) and (v), we getTotal energy of mass -spring system  =      1    2    m      A    2              ω        2        cos    2    ⁡  (      ω    t  +      ϕ    )  +      1    2    k      y    2    +  m  g  x  +      1    2    m            ω        2        A    2        sin    2    ⁡  (      ω    t  +      ϕ    )  −  m  g  x  =      1    2    m      A    2              ω        2    +      1    2    k      y    2

When a mass m is hung from a spring of negligible mass with a force constant k, it extends by an amount y.

Answered question

Answer & Explanation

New Questions in Force, Motion and Energy