Oscillations and Simple Harmonic Motion - class-XI

For the amplitude of oscillation and energy to be maximum, the frequency of force must be equal to the initial frequency and this is only possible in resonance. In resonance state $ \omega _1 = \omega _2$.

When a spring of force constant k is cut into three equal parts, the force constant of each part becomes 3k. When a mass of 3m is suspended from one part, the new frequency is f_new = (1/(2pi)) * sqrt(3k / 3m) = (1/(2pi)) * sqrt(k / m) = f. Thus, the frequency remains unchanged.

For small oscillations, U = 4(1 - cos(2x)) approx 4(1 - (1 - (2x)^2/2)) = 4(2x^2) = 8x^2. The force F = -dU/dx = -16x. Comparing to F = -kx, k = 16. Period T = 2pi * sqrt(m/k) = 2pi * sqrt(1/16) = 2*pi * 1/4 = pi/2.

For a sphere of radius r rolling/sliding in a concave mirror of radius R, the effective length of the pendulum is (R-r). The time period for a simple pendulum is T = 2pi * sqrt(L/g). Substituting L = R-r, we get T = 2pi * sqrt((R-r)/g).

For a damped oscillator, amplitude A(t) = A0 * e^(-bt). Given A(5) = 0.9 * A0, so e^(-5b) = 0.9. In another 10s (total 15s), A(15) = A0 * e^(-15b) = A0 * (e^(-5b))^3 = A0 * (0.9)^3 = 0.729 * A0.

This is a complex problem involving magnetic forces and SHM. The force on the central wire displaced by z is F = - (mu0 * I^2 * z) / (pi * d^2). Using F = ma, where m = lambda * L, the frequency is derived from the restoring force constant.

It is given that $g$ is determined by $:g = \frac{{4{\pi ^2}l}}{{{T^2}}}$

The acceleration is given by a = -beta(x-2). This is the standard form of simple harmonic motion a = -omega^2(x-x0), where omega^2 = beta. The time period T is 2pi/omega, which simplifies to 2pi/sqrt(beta).

A physical pendulum's time period is T = 2pisqrt(I/mgd). Option A describes a physical pendulum where the moment of inertia and distance from the center of mass can be calculated to find the period.

For small oscillations, U(x) is approximated by a quadratic: U(x) approx (1/2)kx^2. Here, U(x) = U0(1 - cos(alphax)) approx U0(alpha^2 * x^2 / 2). Thus k = U0alpha^2. T = 2pisqrt(m/k) = 2pi*sqrt(m / (U0 * alpha^2)).

When a boy stands up on a swing, his center of mass moves upward, effectively shortening the effective length of the pendulum. Since the time period of a simple pendulum is proportional to the square root of its effective length, reducing the length decreases the time period. Therefore, the new time period is less than T.

According to the question, a and b points are such that they are

For a simple pendulum, T = 2pisqrt(l/g), so g = 4pi^2l / T^2. The relative error is dg/g = dl/l + 2dT/T. Thus, the percentage error is e1 + 2e2.

The frequency of oscillation is given by the change in phase divided by 2 * pi * time interval. Here, delta(theta) = 14 * pi and t = 3.5 seconds. Thus, frequency f = (14 * pi) / (2 * pi * 3.5) = 14 / 7 = 2 Hz.

According to question,

$F _1=-K _1\,x\;\&\;F _2=-K _2\,x$

So $n _1=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1}{m}}=6\;Hz;$

$n _2=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _2}{m}}=8\;Hz$

Now $F=F _1+F _2=-(K _1+K _2)x$

Therefore $n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1+K _2}{m}}$

$\Rightarrow n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{4\pi^2\,n^2 _1\,m+4\pi^2\,n^2 _2\,m}{m}}$$=\sqrt{n _1^2+n _2^2}=\sqrt{8^2+6^2}$

$=10\;Hz$

The damped frequency is omega' = sqrt(omega0^2 - gamma^2). The time period T' = 2pi/omega'. For small damping, T' approx T(1 + gamma^2 / (2omega0^2)). With r^2/mk = 0.8, the change is small and negative/positive depending on the exact definition.

In forced vibration, the resonance wave becomes very sharp when damping force is small (i.e. negligible)

$T=2 \pi \sqrt{\dfrac{L}{g}}$               (where L= length of sting)
From above equation, Time period only depend on the length of the string and g.

In damped oscillation like the motion of simple pendulum swinging in
air, motion dies out eventually. This is because of the air drag and the
friction at the support oppose the motion of the pendulum and dissipate
its energy gradually.

In forced oscillations, the frequency of particle after free
oscillations die out, is the frequency of the driving force, not the
natural frequency of the particle.

$F=-4x-8\F=-4(x+2)\ \Rightarrow F=-4[x-(-2)]\F=-kx\rightarrow$ Distance mean position. So, this will be SHM

To maintain contact with the vibrating platform, the maximum downward acceleration of the platform cannot exceed acceleration due to gravity g. The maximum acceleration in simple harmonic motion is given by A * omega^2 = A * (2 * pi * f)^2. Setting this equal to g gives A * (2 * pi * 0.5)^2 = 9.8, which solves to A = 9.8 / (pi^2) approximately equal to 0.9927 meters.