Two oscillating systems; a simple pendulum and a vertical spring-mass-system have same time period of motion on the surface of the Earth. If both are taken to the moon, then-
  • A
    Time period of the simple pendulum will be more than that of the spring-mass system.
  • B
    Time period of the simple pendulum will be equal is that is of the spring-mass system.
  • C
    Time period of the simple pendulum will be less than of the spring-mass system.
  • D
    Nothing can be said definitely without observation.
Medium
art

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Similar Questions

  • 1
    Two damped spring-mass oscillating systems have identical spring constants and decay times. However, system $A's$ mass $m_A$ is twice system $B's$ mass $m_B$ . How do their damping constants, $b$ , compare ?
    View Solution
  • 2
    The potential energy of a particle of mass $0.1\,kg,$ moving along $x-$ axis, is given by $U = 5x(x-4)\,J$ where $x$ is in metres. It can be concluded that
    View Solution
  • 3
    A mass $0.9\,kg$, attached to a horizontal spring, executes $SHM$ with an amplitude $A _{1}$. When this mass passes through its mean position, then a smaller mass of $124\,g$ is placed over it and both masses move together with amplitude $A _{2}$. If the ratio $\frac{ A _{1}}{ A _{2}}$ is $\frac{\alpha}{\alpha-1}$, then the value of $\alpha$ will be$......$
    View Solution
  • 4
    In the given figure, a mass $M$ is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is $k$. The mass oscillates on a frictionless surface with time period $T$ and amplitude $A$. When the mass is in equilibrium position, as shown in the figure, another mass $m$ is gently fixed upon it. The new amplitude of oscillation will be
    View Solution
  • 5
    A point particle of mass $0.1\, kg$ is executing $S.H.M$. of amplitude of $0.1\, m$. When the particle passes through the mean position, its kinetic energy is $8\times10^{-3}$ Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is $45^o$.
    View Solution
  • 6
    A second's pendulum is placed in a space laboratory orbiting around the earth at a height $3R$, where $R$ is the radius of the earth. The time period of the pendulum is
    View Solution
  • 7
    A graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion is
    View Solution
  • 8
    There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4,$ then the new time period of the pendulum is
    View Solution
  • 9
    Time period of a particle executing $SHM$ is $8\, sec.$ At $t = 0$ it is at the mean position. The ratio of the distance covered by the particle in the $1^{st}$ second to the $2^{nd}$ second is :
    View Solution
  • 10
    Abody performs simple harmonic oscillations along the straight line $ABCDE$ with $C$ as the midpoint of $AE.$ Its kinetic energies at $B$ and $D$ are each one fourth of its maximum value. If $AE = 2R,$ the distance between $B$ and $D$ is
    View Solution