$1+\omega\text{t}+\omega^2\text{t}^2$
$\therefore$ The amplitude is $\sqrt{3^2+4^2},$ i.e., 5 units.
$\text{x}=\text{a}\cos\omega\text{t}$
Velocity $\text{v}=\frac{\text{dx}}{\text{dt}}=\text{a}\omega(-\sin\omega\text{t})=-\text{a}\omega\sin\omega\text{t}$$\Rightarrow\text{v}=\text{a}\omega\cos\Big(\frac{\pi}{2}+\omega\text{t}\Big)$
Now, phase of displacement $\phi_1=\omega\text{t}$ Phase of velocity $\phi_2=\frac{\pi}{2}+\omega\text{t}$$\therefore$ Difference in phase of velocity to that of phase of displacement
$\triangle\phi=\phi_2-\phi_1=\Big(\frac{\pi}{2}+\omega\text{t}\Big)-(\omega\text{t})=\frac{\pi}{2}$
$=\frac{8}{9}\frac{1}{2}\text{m}\omega^2\text{A}^2=\frac{8}{9}\text{E}$
$\text{v}_1=\omega\sqrt{\text{A}^2-\text{x}_1^2}$
$\text{v}_1^2=\omega^2(\text{A}^2-\text{x}_1^2)\cdots\text{(i)}$
For displacement $\text{x}=\text{x}_2$$\text{v}_2=\omega\sqrt{\text{A}^2-\text{x}_2^2}$
$\text{v}_2^2=\omega^2(\text{A}^2-\text{x}_2^2)\cdots\text{(ii)}$
Subtracting (i) from (ii), we have$\text{v}_2^2=\text{v}_2\omega^2(\text{x}_1^2-\text{x}_2^2)$
$\Rightarrow\omega=\sqrt{\frac{(\text{v}_2^2-\text{v}_1^2)}{(\text{x}_1^2-\text{x}_2^2)}}$
$\therefore$ Period of oscillation $\text{T}=\frac{2\pi}{\omega}$
$=2\pi\sqrt{\frac{(\text{x}_2^2-\text{x}_2^2)}{(\text{v}_1^2-\text{v}_2^2)}}$
$\because\omega^2=16\Rightarrow\omega=\frac{2\pi}{\text{T}}=\sqrt{16}=4$
$\therefore\text{T}=\frac{\pi}{2}\text{ second}$
$\Rightarrow\text{F = kx}$
$\Rightarrow\text{F}=-\text{m}\omega^2\text{x}.$
$\text{T}=2\pi\sqrt{\frac{\text{l}}{\sqrt{\text{g}^2+\text{a}^2}}}.$
Density of water = 10
3kg When the cylinder is depressed in water through a distance y the restoring force = weight of water displaced F = Aydg = (15 × 10-4) × 103 × 9.8 newton/ metre = 1.5 × 9.8 N/ m Hence the frequenry of orillation is given by$=\frac{1}{2\pi}\sqrt{\Big(\frac{\text{k}}{\text{m}}\Big)}$
$=\frac{1}{2\pi}\sqrt{\frac{1.5\times9.8}{0.28}}$
$=1.15\text{Hz}$
$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
Here, I = 0, so T becomes infinity.$\because\omega^2=16$
$\Rightarrow\omega=\frac{2\pi}{\text{T}}=\sqrt{16}=4$
$\text{T}=\frac{\pi}{2}\text{ sec}.$
Developing a restoring force Kx towards Left on the block. F1 = -Kx (for left spring) and F2 = -Kx (for right spring) Restoring force, F = F1 + F2 =-2Kx $\therefore$ F = 2Kx towards left.
$\text{f}=\frac{1}{2\pi}\sqrt{\frac{\text{k}_1+\text{k}_2}{}\text{m}}$
$\text{l}=\frac{\text{T}^2\text{g}}{4\pi^2}=\frac{4\text{g}}{4\pi^2}=1\text{m}$
General vibrations of a polyatomic molecule about its equilibrium position.