A spring executes $SHM$ with mass of $10\,kg$ attached to it. The force constant of spring is $10\,N/m$.If at any instant its velocity is $40\,cm/sec$, the displacement will be .... $m$ (where amplitude is $0.5\,m$)
Medium
Download our appand 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
- 1The phase (at a time $t$) of a particle in simple harmonic motion tellsView Solution
- 2A particle executes $SHM $ in a line $4\, cm $ long. Its velocity when passing through the centre of line is $12\, cm/s$. The period will be ..... $s$View Solution
- 3A 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 areView Solution
- 4The average speed of the bob of a simple pendulum oscillating with a small amplitude $A$ and time period $T$ isView Solution
- 5$Y = A \sin \left(\omega t +\phi_{0}\right)$ is the time-displacement equation of a SHM. At $t=0$ the displacement of the particle is $Y =\frac{ A }{2}$ and it is moving along negative $x$ -direction. Then the initial phase angle $\phi_{0}$ will be ...... .View Solution
- 6In a seconds pendulum, mass of bob is $30\, gm$. If it is replaced by $90\, gm$ mass. Then its time period will .... $\sec$View Solution
- 7The maximum velocity of a body undergoing $S.H.M$. is $0.2\,m/s$ and its acceleration at $0.1\,m$ from the mean position is $0.4\,m/s^2$. The amplitude of the $S.H.M.$ is .... $m$View Solution
- 8A particle executing $SHM$ of amplitude $4\,cm$ and $T = 4s.$ The time taken by it to move from $+2\,cm$ to $+2\sqrt 3\,cm$ isView Solution
- 9The bob of simple pendulum having length $l$, is displaced from mean position to an angular position $\theta$ with respect to vertical. If it is released, then velocity of bob at lowest positionView Solution
- 10A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ${\omega _0}$ - An external force $F (t)$ proportional to $\cos \omega \,t((\omega \ne {\omega _0})$ is applied to the oscillator. The time displacement of the oscillator will be proportional toView Solution