A spring of force constant $k$ is cut into lengths of ratio $1 : 2 : 3$. They are connected in series and the new force constant is $K’$. Then they are connected in parallel and force constant is $K”$. Then $K’ : K”$ is: (2017-Delhi)
$k \propto 1/L$. The parts have stiffness $6k, 3k, 2k$. In series, $K' = k$. In parallel, $K'' = 6k + 3k + 2k = 11k$. The ratio is $1:11$.
A body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released, it oscillates with a time period of $3 \text{ s}$. When the mass $m$ is increased by $1 \text{ kg}$, the time period of oscillations becomes $5 \text{ s}$. The value of $m$ in kg is: (2016 – II)
$T \propto \sqrt{m} \Rightarrow \frac{3}{5} = \sqrt{\frac{m}{m+1}} \Rightarrow \frac{9}{25} = \frac{m}{m+1} \Rightarrow 25m = 9m + 9 \Rightarrow m = \frac{9}{16} \text{ kg}$.
The period of oscillation of a mass $M$ suspended from a spring of negligible mass is $T$. If along with it another mass $M$ is also suspended, the period of oscillation will now be: (2010 Pre)
$T = 2\pi\sqrt{\frac{M}{k}}$. When another mass $M$ is suspended, total mass is $2M$. New period $T' = 2\pi\sqrt{\frac{2M}{k}} = \sqrt{2}T$.
Two springs of spring constants $k_1$ and $k_2$ are joined in series. The effective spring constant of the combination is given by: (2004)
For springs in series, the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$. Thus $k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$.
The time period of a mass suspended from a spring is $T$. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be: (2003)
Spring constant is inversely proportional to its length. For a one-fourth part, $k' = 4k$. New period $T' = 2\pi\sqrt{\frac{m}{4k}} = \frac{T}{2}$.
The total energy of particle performing S.H.M. depend on: (2001)
The total energy in S.H.M is $E = \frac{1}{2} K a^2$. Thus, it depends only on the force constant $K$ and amplitude $a$.
A linear harmonic oscillator of force constant $2 \times 10^6 \text{ N/m}$ and amplitude $0.01 \text{ m}$ has a total mechanical energy of $160 \text{ J}$. Its (1996)
Max K.E. = $\frac{1}{2} k a^2 = \frac{1}{2} \times 2 \times 10^6 \times (0.01)^2 = 100 \text{ J}$. Total Energy = $160 \text{ J}$. Min P.E. = $160 - 100 = 60 \text{ J}$. Max P.E. = Total Energy = $160 \text{ J}$.
In a simple harmonic motion, when the displacement is one-half of the amplitude, what fraction of the total energy is kinetic? (1995)
$K.E. = \frac{1}{2} k (a^2 - x^2)$. At $x = a/2$, $K.E. = \frac{1}{2} k (a^2 - a^2/4) = \frac{3}{4} (\frac{1}{2} k a^2) = \frac{3}{4} E$.
A body executes simple harmonic motion with an amplitude $A$. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy? (1992)
$P.E. = \frac{1}{4} E \Rightarrow \frac{1}{2} k x^2 = \frac{1}{4} (\frac{1}{2} k A^2) \Rightarrow x^2 = \frac{A^2}{4} \Rightarrow x = \frac{A}{2}$.
Two pendulums of length $121 \text{ cm}$ and $100 \text{ cm}$ start vibrating in phase. At some instant, the two are at their means position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the means position is : (2022)
$T \propto \sqrt{l}$. So $T_1/T_2 = \sqrt{121/100} = 11/10$. This gives $10 T_1 = 11 T_2$. The shorter pendulum ($T_2$) completes $11$ vibrations.