NEET Physics: Waves and Oscillations questions with solutions

Q1. Two waves represented by \( y_{1}= \) \( 10 \sin (2000 \pi t) \) and \( y_{2}= \) \( 10 \sin (2000 \pi t+\pi / 2) \) are superimposed at any point at a particular instant. The resultant amplitude is?

1. 10 units
2. 20 units
3. 14.1 units
4. zero

Answer: 14.1 units

The two waves have the same amplitude, 10, and a phase difference of \(\pi/2\). For equal amplitudes \(A\) with phase difference \(\phi\), the resultant amplitude is \(2A\cos(\phi/2)\). Substituting gives \(2\cdot10\cdot\cos(\pi/4)=20\cdot\frac{\sqrt2}{2}=14.1\).

Q2. The equation of a plane progressive wave is given by \( \boldsymbol{y}=\boldsymbol{A} \sin (\boldsymbol{\omega} \boldsymbol{t}-\boldsymbol{k} \boldsymbol{x}) \) The dimension of \( (\omega / k) \) is that of

1. frequency
2. velocity
3. wavelength
4. inverse of velocity

Answer: velocity

In a wave equation, the quantity inside the sine must be dimensionless, so ω has units of s⁻¹ and k has units of m⁻¹. Therefore ω/k has units of m s⁻¹, which is velocity.

Q3. Identify the parameter which measures the time it takes to complete one cycle?

1. period
2. wavelength
3. frequency
4. amplitude E. speed

Answer: period

The period is the time required for one complete cycle of a repeating motion or wave. Frequency tells how many cycles occur per second, so the time for one cycle is the period.

Q4. A plane progressive wave travelling in \( -Y \) direction is represented by the equation \( 2 \cos (2 \pi t+\pi y) . \) If this wave was travelling in \( X \) direction, the frequency of the wave would have been

1. doubled
2. remains same
3. tripled
4. halved

Answer: remains same

The frequency of the wave remains the same regardless of the direction of travel. [AI-generated key — verify before high-stakes use]

Q5. A vibrator makes \( 150 \mathrm{cm} \) of a string to vibrate in 6 loops in the longitudinal arrangement when it is stretched by 150 N. The entire length of the string is then weighed and is found to weigh \( 400 \mathrm{mg} \) Then This question has multiple correct options

1. Frequency of the vibrator is 3 \( \mathrm{kHz} \)
2. Frequency of the vibrator is \( 1.5 \mathrm{kHz} \)
3. Distance between two nodes is \( 25 \mathrm{cm} \)
4. Distance between two nodes is 33 cm

Answer: Distance between two nodes is \( 25 \mathrm{cm} \)

In a standing wave, each loop is half a wavelength, so 6 loops in 150 cm means one loop is 25 cm. The distance between adjacent nodes is also half a wavelength, so it is 25 cm.