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Worksheet 1 Solutions: Waves
Student Name: ____________________________________ Class: 11th (Physics) Subject: Physics
Worksheet 1 Solutions
1.
Distinguish between Transverse and Longitudinal waves with examples.
Sol: Transverse: Medium particles oscillate perpendicular to wave direction (e.g. Light, waves on string). Longitudinal: Medium particles oscillate parallel to wave direction (e.g. Sound waves).
2.
State Laplace's Correction for speed of sound in air.
Sol: Sound propagation is an adiabatic process, not isothermal: $v = \sqrt{\frac{\gamma P}{\rho}}$, where $\gamma = 1.41$ for air.
3.
What are standing (stationary) waves? State nodes and antinodes.
Sol: Formed by superposition of two identical waves travelling in opposite directions. Nodes: Zero amplitude points; Antinodes: Maximum amplitude points.
4.
State Doppler Effect formula for sound source moving towards a stationary observer.
Sol: Apparent frequency $f' = f \left(\frac{v}{v - v_s}\right)$.
5.
A progressive wave equation is $y = 0.05 \sin(20\pi t - 2\pi x)\text{ m}$. Find amplitude, frequency, wavelength, and wave speed.
Sol: $A = 0.05 \text{ m}$, $\omega = 20\pi \implies f = 10 \text{ Hz}$, $k = 2\pi \implies \lambda = 1 \text{ m}$, $v = f \lambda = 10 \text{ m/s}$.
6.
A stretched string of length $1 \text{ m}$ and mass $10 \text{ g}$ is under tension $100 \text{ N}$. Find fundamental frequency.
Sol: Linear mass density $\mu = 0.01 \text{ kg/m}$. $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.01}} = 100 \text{ m/s}$. $f_1 = \frac{v}{2L} = \frac{100}{2(1)} = 50 \text{ Hz}$.
7.
An organ pipe open at both ends has fundamental frequency $250 \text{ Hz}$. Find fundamental frequency if one end is closed.
Sol: $f_{open} = \frac{v}{2L} = 250 \text{ Hz}$. $f_{closed} = \frac{v}{4L} = \frac{f_{open}}{2} = 125 \text{ Hz}$.
8.
Two tuning forks of frequencies $256 \text{ Hz}$ and $260 \text{ Hz}$ are sounded together. Find beat frequency and time interval between beats.
Sol: Beat frequency $f_b = |260 - 256| = 4 \text{ Hz}$. Time interval $T_b = \frac{1}{f_b} = \frac{1}{4} = 0.25 \text{ seconds}$.