Physics 3 Marks Question

$=2\cos\Big(3\text{t}+\frac{5\pi}{6}\Big)$

If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:

Amplitude, A = 2cm

Phase angle, $\phi=\frac{5\pi}{6}=150^\circ$

Angular velocity, $\omega=\frac{2\pi}{\text{T}}=3\text{rad/sec.}$

The motion of the particle can be plotted as shown in the following figure.

If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:

The motion of the particle can be plotted as shown in the following figure.

$=-3\cos\bigg[\Big(2\pi\text{t}+\frac{\pi}{4}\Big)+\frac{\pi}{2}\bigg]=-3\cos\Big(2\pi\text{t}+\frac{3\pi}{4}\Big)$

If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:

Amplitude, A = 3cm

Phase angle, $\phi=\frac{3\pi}{4}=135^\circ$

Angular velocity, $\omega=\frac{2\pi}{\text{T}}=2\pi\text{ rad/s}$

The motion of the particle can be plotted as shown in the following figure.

If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:

Amplitude, A = 1

Phase angle, $\phi=-\frac{\pi}{6}=-30^\circ$

Angular velocity, $\omega=\frac{2\pi}{\text{T}}=1\text{rad/s}$

The motion of the particle can be plotted as shown in the following figure.

1. At t = 0

$\text{x}_1=\text{A}_1\sin\omega\text{t}=0$

$\text{x}_2=\text{A}_2\Big(\omega+\frac{\pi}{3}\Big)$

$=\frac{\text{A}_2\sqrt{3}}{2}$

Thus the resultant displacement at t = 0 is

$\text{x}=\text{x}_1+\text{x}_2=\frac{\text{A}_2\sqrt{3}}{2}$

2. $\text{A}=\sqrt{\text{A}_1^2+{\text{A}_2^2}+2\text{A}_1\text{A}_2^2\cos\frac{\pi}{3}}$

$\text{A}=\sqrt{\text{A}_1^2+{\text{A}_2^2}+\text{A}_1\text{A}_2}$

The maximum speed is

$\text{V}_\text{max}=\omega\text{A}$

$=\omega=\sqrt{\text{A}_1^2+{\text{A}_2^2}+\text{A}_1\text{A}_2}$

3. The maximum acceleration is

$\text{a}_\text{max}=\omega^2\text{A}$

$=\omega^2=\sqrt{\text{A}_1^2+{\text{A}_2^2}+\text{A}_1\text{A}_2}$

Let the mass be displaced by a small distance x to the right of equilibrium position. Due to this, spring on left gets elongated by length equal to x and that on the right side gets compressed by same length. Then force acting on masses are

F₁ = -kx (force acting on left side and trying to pull the mass towards the mean position.)

F₂ = -kx (force exerted by spring on right side trying to push the mass towards mean position.)

Net force F, acting on the mass

F = -2kx

$\therefore$ Force acting on mass is directly propotional to displacement and it directed towards mean position.

$\therefore$ Motion is simple harmonic and time period of oscillation is

$\text{T}=2\pi\sqrt{\frac{\text{m}}{2\text{k}}}$

1. Force constant $\text{k}=\frac{\text{F}}{\text{l}}=\frac{\text{mg}}{\text{l}}$

Here m = 3.0kg and elongation in length of spring l = 0.2m

$\therefore$ Force constant $\text{k}=\frac{3.0\times9.8}{0.2}=174\text{Nm}^{-1}$

2. Period of oscillation $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$

$=2\times3.14\times\sqrt{\frac{3}{147}}=0.9\text{s}$

1. When the springs are connected in parallel, the extension in them will be same and the total restoring force is the sum of their restoring forces.

$\therefore\text{F = F}_1+\text{F}_2$

$-\text{k}_{\text{eq}}\text{x}=-\text{k}_1\text{x}-\text{k}_2\text{x}$

$\text{k}_{\text{eq}}=\text{k}_1+\text{k}_2$

2. When the springs are connected in series, the restoring force is same in both the springs and the extensions will be different so the not extension

i.e., $\text{x = x}_1+\text{x}_2$

$=\frac{\text{F}}{-\text{k}_{\text{eq}}}=\frac{-\text{F}}{\text{k}_1}-\frac{\text{F}}{\text{k}_2}$

$\therefore\frac{1}{\text{k}_{\text{eq}}}=\frac{1}{\text{k}_1}+\frac{1}{\text{k}_2}$

when connected in series.

1. Motion is always directed towards a fixed point or equilibrium point.
2. Motion being represented by bounded trigonometric functions.
3. Acceleration is directly proportional to negative of displacement, i.e., $\text{a}\propto-\text{x}$

Equation for S.H.M.

Acceleration $=-\omega^2\text{x}$

$\Rightarrow\frac{\text{d}^2\text{x}}{\text{dt}^2}+\omega^2\text{x}=0,\omega=2\pi\text{f}$

$\omega$ is angular frequency (radian/ sec), f is linear fiequency (s^-1) or (hertz)

$\text{F}=-\text{mg}\sin\theta$

$\theta=$ Angle of displacement

1. The sound heard by an observer/ listner after the reflection from a surface is called echo.
2. Total distance travelled by sound to come back = 2 × 860 = 1720m speed of sound is 344m/ s.

$\therefore\text{Time}=\frac{\text{Distance}}{\text{Speed}}$

$=\frac{1720}{344}=5\text{sec}$

$\text{T}=\frac{1}{2}\text{m}\omega^2(\text{A}^2-\text{x}^2)$

$\text{V}=\frac{1}{2}\text{m}\omega^2\text{x}^2$

$\frac{\text{T}}{\text{V}}=\frac{\text{A}^2-\text{x}^2}{\text{x}^2}=\Big(\frac{\text{A}^2}{\text{x}^2}-1\Big)$

$\Big(\frac{2\pi}{\text{T}}\text{t}+\phi\Big)$

$\text{x = a}\cos\omega\text{t}$

$\frac{\text{dx}}{\text{dt}}=\omega\text{a}\sin\omega\text{t}$

$\therefore\text{Velocity},\text{v}=\frac{\text{dx}}{\text{dt}}$ or $\text{v}^2=\text{a}^2\omega^2\sin^2\omega\text{t}.$

$=\text{a}^2\omega^2(1-\cos^2)\omega\text{t}$

$=\omega^2(\text{a}^2-\text{x}^2)$

$=\frac{1}2{}\text{Kx}^2=\frac{1}{2}\text{m}\omega^2\text{x}^2$

$=\frac{1}2{}\text{mv}^2=\frac{1}2{}\text{m}\big[\omega\sqrt{\text{A}^2-\text{x}^2}\big]^2$

$=\frac{1}{2}\text{m}\omega^2(\text{A}^2-\text{x}^2)$

$\therefore\text{m}=\frac{1}{6\times10^{26}},\text{v}=9\times10^{13}\text{s}^{-1}$

$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$

$\text{v}^2=\frac{1}{4\pi^2}\times\frac{\text{k}}{\text{m}}$

$\text{k}=4\pi^2\text{v}^2\text{m}$

$=4(3.14)^2(9\times10^{13})\times\frac{1}{6\times10^{26}}\text{Nm}^{-1}$

$=5.32\times10^2\text{Nm}^{-1}$

1. Time period is given by

$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{5}{100}}=0.41\text{s}$

2. Angular velocity,

$\omega=\frac{2\pi}{\text{T}}=\frac{2\pi}{0.41}=4.5\text{ rad/ s}$

Maximum $\text{K.E.}=\text{E}_{\text{Kmax}}=\frac{1}{2}\text{mv}^2_0$

$=\frac{1}2{}\text{m}(\omega\text{A})^2=12.50\text{J}$

Maximum $\text{P.E. = E}_{\text{Pmax}}$

$=\frac{1}{2}\text{k}\text{A}^2=12.50\text{J}$

Total energy $\text{E = E}_{\text{Kmax}}$

$=\text{E}_{\text{Pmax}}=12.50\text{J}$

$\text{PE}=\frac{1}{2}\text{kx}^2=\frac{1}{2}\text{m}\omega^2\text{x}^2$

$\text{TE}=\frac{1}{2}\text{m}\omega^2\text{A}^2$

$\text{a}_{\text{max}}=\omega^2\text{A}=(2\pi\text{v})^2\text{A}=4\pi^2\text{v}^2\text{A}$

$=4\times\Big(\frac{2}{7}\Big)^2\times(2)^2\times0.05$

$=7.9\text{ms}^{-2}$

$\therefore$ Maximum force felt by the man

$\therefore\text{ma}=-\text{kx,a}=\frac{-\text{k}}{\text{m}}\text{x}$