A vertical mass spring system executes simple harmonic oscillations with a period of $2\,s$. A quantity of this system which exhibits simple harmonic variation with a period of $1\, sec$ is
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A particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x)-\frac{1}{2} k x^2-V_0 \cos \left(\frac{x}{a}\right)$, where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$, the time period is given by
Kinetic energy of a particle executing simple harmonic motion in straight line is $pv^2$ and potential energy is $qx^2$, where $v$ is speed at distance $x$ from the mean position. It time period is given by the expression
The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\, cm$ to $8\, cm$ in $40\, seconds$ . Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$ . The time in which amplitude of this pendulum will reduce from $10\, cm$ to $5\, cm$ in carbon dioxide will be close to ..... $s$ $(ln\, 5 = 1.601,ln\, 2 = 0 .693)$
The length of a seconds pendulum at a height $h=2 R$ from earth surface will be.(Given: $R =$ Radius of earth and acceleration due to gravity at the surface of earth $g =\pi^{2}\,m / s ^{-2}$ )
A flat horizontal board moves up and down in $SHM$ of amplitude $\alpha$. Then the shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is
A body executing $S.H.M.$ along a straightline has a velocity of $3 \,ms ^{-1}$ when it is at a distance of $4 \,m$ from its mean position and $4 \,ms ^{-1}$ when it is at a distance of $3 \,m$ from its mean position. Its angular frequency and amplitude are
A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of time period $T$. If the mass is increased by m, the time period becomes $5T/3$. Then the ratio of $m/M$ is