Physics 3 Marks Question

$\text{m}_1=\text{m}_2=3.2\text{kg}$

mass per unit length of $\text{AB}=10\text{g/mt}=0.01\text{kg/mt}$

mass per unit length of $\text{CD}=8\text{g/mt}=0.008\text{kg/mt}$

for the string CD, $\text{T}=3.2\times\text{g}$

$\Rightarrow\text{v}=\sqrt{\Big(\frac{\text{T}}{\text{m}}\Big)}=\sqrt{\Big(\frac{3.2\times10}{0.008}\Big)}=\sqrt{\frac{(32\times10^3)}{8}}\\=2\times10\sqrt{10}=20\times3.14=63\text{m/s}$

for the string AB, $\text{T}=2\times3.2\text{g}=6.4\times\text{g}=64\text{N}$

$\Rightarrow\text{v}=\sqrt{\Big(\frac{\text{T}}{\text{m}}\Big)}=\sqrt{\Big(\frac{64}{0.01}\Big)}$

$=\sqrt{6400}$

$=80\text{m/s}$

1. Velocity of the wave, $\text{V}=\sqrt{\frac{\text{T}}{\text{m}}}$

$=\sqrt{\frac{(16\times10^5)}{0.4}}=2000\text{cm/sec}$

$\therefore\ $Time taken to reach to the other end $=\frac{20}{2000}=0.01\text{sec}$

Time taken to see the pulse again in the original position = 0.01 ×2 = 0.02sec.

2. At t = 0.01s, there will be a ‘though’ at the right end as it is reflected.

$\text{V}=200\text{m/s},\ 2\text{A}=0.5\text{m}$

1. The string is vibrating in its 1^st overtone

$\Rightarrow\lambda=1=2\text{m}$

$\Rightarrow\text{f}=\frac{\text{v}}{\lambda}=100\text{Hz}$

2. The stationary wave equation is given by

$\text{y}=2\text{A}\cos\frac{2\pi\text{x}}{\lambda}\sin\frac{2\pi\text{Vt}}{\lambda}$

$=(0.5\text{cm})\cos\big[(\pi\text{m}^{-1})\text{x}\big]\sin\big[(200\pi\text{s}^{-1})\text{t}\big]$

$\text{L}=40\text{cm}=0.4\text{m},\ \text{mass}=3.2\text{kg}=3.2\times10^{-3}\text{kg}$

$\therefore\ $mass per unit length, $\text{m}=\frac{(3.2)}{(0.4)}=8\times10^{-3}\text{kg/m}$

Change in length, $\triangle\text{L}=40.05-40=0.05\times10^{-2}\text{m}$

strain $=\frac{\triangle\text{L}}{\text{L}}=0.125\times10^{-2}\text{m}$

$\text{f}=220\text{Hz}$

$\text{f}=\frac{1}{2\text{l}'}\sqrt{\frac{\text{T}}{\text{m}}}=\frac{1}{2\times(0.4005)}\sqrt{\frac{\text{T}}{8\times10^{-3}}}$

$\Rightarrow\text{T}=248.19\text{N}$

Strain $=\frac{248.19}{1}\text{mm}^2=248.19\times10^6$

$\text{Y}=\frac{\text{Stress}}{\text{Strain}}=1.985\times10^{11}\text{N/m}^2$

$\text{y}=(0.10\text{mm})\sin\big[(31.4\text{m}^{-1})\text{x}+(314\text{s}^{-1})\text{t}\big].$ $\Big(\text{y}=\text{r}\sin\Big\{\big(\frac{2\pi\text{x}}{\lambda}\big)\Big\}+\omega\text{t}\Big)$

1. Negative x-direction

2. k = 31.4m^-1

$\Rightarrow\frac{2\lambda}{\lambda}=31.4$

$\Rightarrow\lambda=\frac{2\pi}{31.4}=0.2\text{mt}=20\text{cm}$

Again, $\omega=314\text{s}^{-1}$

$\Rightarrow2\pi\text{f}=314$

$\Rightarrow\text{f}=\frac{314}{2\pi}=\frac{314}{\big(2\times\frac{3}{14}\big)}=50\text{sec}^{-1}$

$\therefore\ $wave speed, $\text{v}=\lambda\text{f}=20\times50=1000\text{cm/s}$

3. Max. displacement = 0.10mm

Max. velocity $=\text{a}\omega=0.1\times10^{-1}\times314=3.14\text{cm/sec}$

$\text{y}(0.4\text{cm})\sin\big[(0.314\text{cm}^{-1})\text{x}\big]\cos\big[(600\pi\text{s}^{-1})\text{t}\big]$

1. $\omega=600\pi\Rightarrow2\pi\text{f}=600\pi\Rightarrow\text{f}=300\text{Hz}$

Wavelength,, $\lambda=\frac{2\pi}{0.314}=\frac{(2\times3.14)}{0.314}=20\text{cm}$

2. Therefore nodes are located at, $0,\ 10\text{cm},\ 20\text{cm},\ 30\text{cm}$

3. Length of the string $=\frac{3\lambda}{2}=3\times\frac{20}{2}=30\text{cm}$

4. $\text{y}=0.4\sin(0.314\text{x})\cos(600\pi\text{t})$

$\Rightarrow0.4\sin\Big\{\big(\frac{\pi}{10}\big)\text{x}\Big\}\cos(600\pi\text{t})$

since $\lambda$ and v are the wavelength and velocity of the waves that interfere to give this vibration $\lambda=20\text{cm}$

$\text{v}=\frac{\omega}{\text{k}}=6000\text{cm/sec}=60\text{m/s}$

$\rho_\text{s}=7.8\text{g/cm}^3,\ \rho_\text{A}=2.6\text{g/cm}^3$

$\text{m}_\text{s}=\rho_\text{s}\text{A}_\text{s}=7.8\times10^{-2}\text{g/cm}$ (m = mass per unit length)

$\text{m}_\text{A}=\rho_\text{A}\text{A}_\text{A}=2.6\times10^{-2}\times3\text{g/cm}$

$=7.8\times10^{-3}\text{kg/m}$

$\Rightarrow\text{v}=\sqrt{\frac{\text{T}}{\text{m}}}$

$\Rightarrow\frac{500}{7}\text{m/x}$

$\therefore\ $maximum distance of a loop $=20\text{cm}$

$\Rightarrow\text{v}_1=\sqrt{\Big(\frac{\text{T}}{\text{m}_1}\Big)}$

$\text{v}=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\text{m}}}$

$\Rightarrow\sqrt{\frac{\text{T}}{\text{m}}}=\text{v}2\text{l}$ $\Big[\sqrt{\frac{\text{T}}{\text{m}}}=$ velocity of wave$\Big]$

1. Wavelength, $\lambda=\frac{\text{velocity}}{\text{frequency}}=\frac{\text{v}2\text{l}}{\text{v}}=2\text{l}$

and wave number $\text{K}=\frac{2\pi}{\lambda}=\frac{2\pi}{2\text{l}}=\frac{\pi}{\text{l}}$

2. Therefore, equation of the stationary wave is

$\text{y}=\text{A}\cos\Big(\frac{2\pi\text{x}}{\lambda}\Big)\sin\Big(\frac{2\pi\text{Vt}}{\text{L}}\Big)$

$=\text{A}\cos\Big(\frac{2\pi\text{x}}{2}\Big)\sin\Big(\frac{2\pi\text{Vt}}{\text{2L}}\Big)$

$\text{v}=\frac{\text{V}}{2\text{L}}$ $\Big[$because $\text{v}=\big(\frac{\text{v}}{2\text{l}}\big)\Big]$

1. To again original shape distance travelled by the wave S = 20 + 20 = 40cm.

Wave speed, v = 20m/s

$\Rightarrow\text{Times}=\frac{\text{s}}{\text{v}}$

$=\frac{40}{20}$

$=2\text{sec}$

2. The wave regains its shape, after traveling a periodic distance = 2 × 30 = 60cm

$\therefore\ $Time period $=\frac{60}{20}=3\text{sec}.$

3. Frequency, $\text{n}=\Big(\frac{1}{3}=3\text{sec}^{-1}\Big)$

$\text{n}=\Big(\frac{1}{2\text{l}}\Big)\sqrt{\frac{\text{T}}{\text{m}}}$

m = mass per unit length = 0.5g/cm

$\Rightarrow\frac{1}{3}=\frac{1}{(2\times30)}\sqrt{\Big(\frac{\text{T}}{0.5}\Big)}$

$\Rightarrow\text{T}=400\times0.5=200\ \text{dyne}$

$=2\times10^{-3}\ \text{Newton}$

$\therefore\ $Mass per unit length = $\text{m}=\frac{5}{64}\text{g/cm}$

$\therefore\ $Tension, T = 8N = 8 × 10⁵ dyne

$\text{V}=\sqrt{\frac{\text{T}}{\text{m}}}$

$=\sqrt{\frac{(8\times10^5\times64)}{5}}$

$=3200\text{cm/s}$

$=32\text{m/s}$