Resonance - class-XI

For a resonance tube, the resonance condition is L + e = (2n-1) * lambda / 4. Using the first two resonances, L1 + e = lambda / 4 and L2 + e = 3 * lambda / 4, we find 2 * (L2 - L1) = lambda. Substituting the values, lambda = 2 * (100 - 32) = 136 cm, so L1 + e = 136 / 4 = 34 cm; thus, e = 34 - 32 = 2 cm.

Amplitude resonance occurs when the driving frequency p equals the natural frequency omega_0 in an undamped system, or slightly below it in a damped system. However, in standard textbook contexts for forced oscillations, p = omega_0 is the condition for amplitude resonance.

$Answer:-$ A,B

The applied force prevents the amplitude from becoming too large.

$Answer:-$ A

During earthquake, when the frequency of earthquake becomes equal to the natural frequency of building, resonance (mechanical) occurs due to which amplitude of vibration of building increases and building get destroy.

$Answer:-$ D

$Aswer:-$ A,B,C

When we push a person in a swing, we generally match the time between two consecutive pushes, with the time, in which swing comes back to its extreme position. In fact we match our pushing frequency with the frequency of swing to get large arc of swing (amplitude), this is a common example of mechanical resonance.

Let $y _1=a\sin \left(\omega t-\cfrac {\pi}{4}\right)$
$y _2=a\sin (\omega t)$
$y _3=a\sin \left(\omega t+\cfrac {\pi}{4}\right)$
On super imposing, resulting SHM-
$y=a\left[\sin \left(\omega t-\cfrac{\pi}{4}\right)+\sin \omega t+\sin \left (\omega t+\cfrac {\pi}{4} \right)\right]$
$\implies y=a \left[2\sin \omega t\cos \cfrac {\pi}{4}+\sin \omega t\right]$
$\implies y=a(1+\sqrt {2})\sin \omega t$
$\therefore$ Resultant amplitude $=(1+\sqrt{2})a$
Also, $\cfrac {E _{resultant}}{E _{single}}=\left(\cfrac {A}{a}\right)^2$
$\implies \cfrac {E _{resultant}}{E _{single}}=(\sqrt{2}+1)^2$
$\implies \cfrac {E _{resultant}}{E _{single}}=(3+2\sqrt{2})$
$\therefore E _{resultant}=(3+2\sqrt{2})E _{single}$

The amplitude of a damped oscillator follows A(t) = A0 * exp(-bt). Given A(2) = A0 / 3, we have exp(-2b) = 1/3. For t = 6, A(6) = A0 * exp(-6b) = A0 * (exp(-2b))^3 = A0 * (1/3)^3 = A0 / 27. Thus, n = 27 = 3^3.

A cylindrical tube,open at one end and closed at the other,is acoustic unison with an external source of frequency held at the open end of the tube, in its fundamental note, then the displacement wave from the source gets reflected with a phase change of  $\pi $ at the closed end.

The cause of rattling sound is resonance and it can be stopped by changing the speed of the vehicle. 

 When the  external frequency becomes equal to the natural frequency of the body , then the amplitude of vibrations of the body becomes very large , this phenomenon is called resonance .

This happens due to resonance (A phenomenon in which a vibrating system or external force drives another system to oscillate with a greater amplitude at a specific frequency). The lamp on top of the piano has a resonant frequency equal to one of the notes being played.

System to absorb more energy when the frequency of its oscillations matches the system's natural frequency of vibration than it does at other frequencies.

Wavelength of the sound produced is,
$\lambda = \dfrac{Velocity}{Frequency} $