Physics 3 Marks Question

Mass of the string, M = 2.50kg

Tension in the string, T = 200N

Length of the string, 1 = 20.0m

Mass per unit length, $\mu=\frac{\text{M}}{\text{L}}=\frac{2.50}{20}=0.125\text{kg m}^{-1}$

The velocity (v) of the transverse wave in the string is given by the relation:

$\text{v}=\sqrt{\frac{\text{T}}{\mu}}$

$=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40\text{m/s}$

Time taken by the disturbance to reach the other end, t = l/v = 20/40 = 0.50s

Also if t₂ is the time taken by the sound to reach at a height h, then

$\text{t}_2=\frac{\text{h}}{\text{v}}=\frac{300}{340}=0.88\text{s}$

$\therefore$ Total time after which sound of splash is heard = t₁ + t₂

$=7.82+0.88=8.7\text{s}$

1. $\text{V}=100 \text{kHz}$

$=100\times10^3\text{Hz},$

$\nu=340\text{ms}^{-1}$

But $\lambda=\frac{\nu}{\text{V}}=\frac{330}{100\times10^3}=3.30\times10^{-3}\text{m}$

2. $\nu$ in water = 1486ms^-1

$\therefore \lambda=\frac{\nu}{\text{V}}=\frac{1486}{10^5}$

$=1.486\times10^{-2}\text{m}$

1. Any wave function should be represented by position and time. y = f(x, t).
2. It should be periodic and simple harmonic. Considering the wave to move with a velocity v, the function can be represented by $\text{x}\mp\nu\text{t}'-'$ sign refers to the motion along +ve x-axis and '+' sign refers to the motion along -ve x-axis.

Mass of the string, M = 2.50kg

Tension in the string, T = 200N

Length of the string, 1 = 20.0m

Mass per unit length, $\mu=\frac{\text{M}}{\text{L}}=\frac{2.50}{20}=0.125\text{kg m}^{-1}$

The velocity (v) of the transverse wave in the string is given by the relation:

$\text{v}=\sqrt{\frac{\text{T}}{\mu}}$

$=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40\text{m/s}$

Time taken by the disturbance to reach the other end, t = l/v = 20/40 = 0.50s

1. $\text{v}'=\text{v}\Big(\frac{\nu}{\nu-\nu_\text{s}}\Big)=1200\Big(\frac{340}{336}\Big)\text{ft}$

$=1214.28\text{Hz}$

2. $\text{v}'=\text{v}\Big(\frac{\nu+\nu_0}{\nu_0}\Big)=1200\Big(\frac{344}{340}\Big)$

$=1214.11\text{Hz}$

1. Travelling wave speed $=\frac{\omega}{\text{k}}=\frac{36}{0.018}$

$=\frac{36}{18}\times10^3$

$=2\times10^3\text{cms}^{-1}$

It travels along the negatice x-axis or from right of left.

2. Amplitude = 3cm

Frequency $=\frac{\omega}{2\pi}=\frac{36}{2\pi}$

$=\frac{18}{\pi}\text{Hz}$

3. Initial phase $=\frac{\pi}{4}$
4. Distabce betweeb successive crests $=\lambda =\frac{2\pi}{\text{k}}$

$=\frac{2\pi}{0.018}=3.5\text{m}$

1. Determine an unknown frequency by listening to the beat frequency $\Delta \text{v}.$ Then unknown frequency $\text{v}'=\text{v}+\Delta\text{v}$ where v is known and it is close to the unknown frequency. Exact values of v is found by loading and filling the turning fork of unknown frequency from which + or - sign is chosen;
2. Tune musical instruments by sounding them together and reducing beats number to zero.
3. Make a sound rich in musical effect by deliberate introduction of beats between different musical instruments.
4. To produce very low frequency pulses which otherwise cannot be produced. Beat frequency is the low frequency sound.
5. Receive radio programme by superheterodyne method.
6. Detect harmful gases in mines.

1. Here nv = 280Hz

and (n + 1)v = 350Hz

$\therefore$ (n + 1)v - nv

= 350 - 280 = 70

v = 70Hz

2. Here $\frac{\text{a}_2}{\text{a}_1}=3$

$\therefore \frac{\text{I}_2}{\text{I}_1}=\Big(\frac{\text{a}_2}{\text{a}_1}\Big)^2=9$

Now, $\log_{10}\Big(\frac{\text{I}_2}{\text{I}_1}\Big)=10\log_{10}\Big(\frac{\text{I}_2}{\text{I}_1}\Big)$

$=10\log_{10}(9)=10\log_{10}3^2\\=20\log_{10}3=\log_{10}3^{20}$

$\therefore \frac{\text{I}_2}{\text{I}_1}=3^{20}$

$\text{I}_2=3^{20}\text{I}_1.$

1. For moist air, values of both λ and ρ are less than the corresponding values for dry air.

$\therefore$ speed of sound in moist air tends to increases due to effect of density $\Big(\text{v}\propto\frac{1}{\sqrt{\rho}}\Big)$ but tends to decrease due to effect of $\lambda (\text{v}\propto\sqrt{\lambda}).$

However the effect of is more than that of $\lambda.$ Hence speed of sound increases with humidity.

2. Speed of sound increases with increases of tem perature $(\text{v}\propto\sqrt{\text{T}}).$

1. The phase difference between two points at a distance one wavelength apart is $2\pi.$ Phase difference between points 0.4m apart is given by:

$=\frac{2\pi}{\lambda}\times0.4=\frac{2\pi}{1.2}\times0.4=\frac{2\pi}{3}\text{radians}$

2. The equation of motion of a progressive wave is:

$\text{y}=\text{A}\sin2\pi\Big(\frac{\text{t}}{\text{T}}-\frac{\text{x}}{\lambda}\Big)$

In the case given

$\text{y}=0.02\sin2\pi\Big(300\text{t}-\frac{\text{x}}{1.2}\Big)$

3. The equation of the stationary wave is:

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

Here $2\text{A}=2\times0.01=0.02$

$\lambda=1.2\text{m}$

$\frac{1}{\text{T}}=300\text{Hz}$

$\therefore \text{y}=0.02\cos\frac{2\pi\text{x}}{1.2}\sin600\pi\text{t}$

4. The distance between the two consecutive nodes in the stationary wave is given by:

$\frac{\lambda}{2}=\frac{1.2}{2}\text{m}=0.6\text{m}$

1. Amplitude = 10 units.
2. Velocity of wave $=\frac{\omega}{\text{k}}=\frac{314}{157}=2\text{ units}.$

1. Amplitude wave $\text{A}=2.2\text{mm}=2.23\times10^{-3}\text{m}$

2. $\omega=300\text{rad s}^{-1}$

$\therefore$ Frequency $\text{v}=\frac{\omega}{2\pi}=\frac{300}{2\pi}=47.7\text{Hz}$

3. $\text{k}=0.24\text{m}^{-1}$

$\therefore$ wavetength $\lambda=\frac{2\pi}{\text{K}}=\frac{2\pi}{0.24}=26.2\text{m}$

4. $\therefore$ Wave velocity $v=\text{v}\lambda=47.7\times26.2=1250\text{ms}^{-1}$

5. Amplitude of particle velocity $=\text{A}\omega=2.2\times10^{-3}\times300=0.66\text{mm}.$

Also if t₂ is the time taken by the sound to reach at a height h, then

$\text{t}_2=\frac{\text{h}}{\text{v}}=\frac{300}{340}=0.88\text{s}$

$\therefore$ Total time after which sound of splash is heard = t₁ + t₂

$=7.82+0.88=8.7\text{s}$