Assuming all pulleys, springs and string massless. Consider all surface smooth. Choose the correct statement $(s)$
Advanced
Download our app for free and get started
$x=\frac{y_{1}}{2}+2 y_{2}+2 y_{3}$ $...(1)$
$\Delta \mathrm{T}=\mathrm{ma}$ $...(2)$
$2 \Delta \mathrm{T}=\mathrm{Ky}_{2}$ $...(3)$
$2 \Delta \mathrm{T}=\mathrm{Ky}_{3}$ $...(4)$
$\frac{\Delta \mathrm{T}}{2}=\mathrm{Ky}_{1}$ $...(5)$
Solving eqn.
$x=\Delta T\left(\frac{33}{4 K}\right)$
$w^{2}=\left(\frac{4 K}{33}\right)$
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
- 1View SolutionThe mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (If it is a second's pendulum on earth)
- 2To find the spring constant $(k)$ of a spring experimentally, a student commits $2 \%$ positive error in the measurement of time and $1 \%$ negative error in measurement of mass. The percentage error in determining value of $\mathrm{k}$ is :View Solution
- 3A tunnel is dug in the earth across one of its diameter. Two masses $‘m’\,\& \,‘2m’$ are dropped from the ends of the tunnel. The masses collide and stick to each other and perform $S.H.M.$ Then amplitude of $S.H.M.$ will be : [$R =$ radius of the earth]View Solution
- 4A $LCR$ circuit behaves like a damped harmonic oscillator. Comparing it with a physical springmass damped oscillator having damping constant $\mathrm{b}$, the correct equivalence would be:View Solution
- 5Two equations of two $S.H.M.$ are $y = a\sin \,(\omega \,t - \alpha )$ and $y = b\cos (\omega \,t - \alpha )$. The phase difference between the two is .... $^o$View Solution
- 6The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ isView Solution
- 7The vertical extension in a light spring by a weight of $1\, kg$ suspended from the wire is $9.8\, cm$. The period of oscillationView Solution
- 8A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A.$ Its speed as it passes through the equilibrium position is $V.$ If extended $2A$ and released, the speed of the mass passing through the equilibrium position will beView Solution
- 9The displacement of a particle undergoing $SHM$ of time period $T$ is given by $x(t) = x_m\,cos\, (\omega t + \phi )$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = + x_m$ whenView Solution
- 10The potential energy of a particle $\left(U_x\right)$ executing $S.H.M$. is given byView Solution