Beats — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Beats MCQs with step-by-step solutions (20 questions). Part of Superposition of Waves. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Beats · easy · theory
Beats are produced by the superposition of two sound waves of:
A. Slightly different frequencies travelling in the same direction ✓ Correct
B. Mutually perpendicular vibrations
C. Widely different frequencies
D. Exactly equal frequencies travelling in opposite directions
Solution: The two waves drift in and out of step, so the resultant amplitude waxes and wanes at the difference frequency.
Q2 — Beats · easy · theory
The beat frequency produced by two sources of frequencies $n_1$ and $n_2$ is:
A. $\dfrac{n_1 + n_2}{2}$
B. $|n_1 - n_2|$ ✓ Correct
C. $n_1 + n_2$
D. $\sqrt{n_1n_2}$
Solution: One beat is heard each time the waves return to being in step, which happens $|n_1 - n_2|$ times per second.
Q3 — Beats · medium · theory
Beats can be distinctly heard only if the beat frequency is:
A. Exactly equal to the frequency of either source
B. Greater than $100\text{ Hz}$
C. Less than about $10\text{ Hz}$ ✓ Correct
D. Zero
Solution: The persistence of hearing is roughly one-tenth of a second, so faster intensity variations merge into a continuous sound.
Q4 — Beats · easy · theory
Beats are used practically to:
A. Measure atmospheric pressure
B. Tune musical instruments to a standard frequency ✓ Correct
C. Determine the density of a gas
D. Measure the speed of light
Solution: As the instrument approaches the reference pitch the beats slow down, vanishing entirely when the frequencies match.
Q5 — Beats · medium · theory
Loading the prong of a tuning fork with a little wax:
A. Stops it vibrating altogether
B. Lowers its frequency, since the vibrating mass increases ✓ Correct
C. Raises its frequency
D. Leaves its frequency unchanged
Solution: Adding mass increases the inertia of the prong, and since $n \propto \dfrac{1}{\sqrt{m}}$ the pitch falls.
Q6 — Beats · medium · theory
Filing the prongs of a tuning fork:
A. Leaves its frequency unchanged
B. Lowers its frequency
C. Makes the fork produce beats by itself
D. Raises its frequency, since the vibrating mass decreases ✓ Correct
Solution: Removing material reduces the inertia of the prong, raising the natural frequency.
Q7 — Beats · easy · theory
The time interval between two successive beats is called the beat period, which equals:
A. The reciprocal of the beat frequency ✓ Correct
B. The beat frequency itself
C. The sum of the two periods
D. The average of the two periods
Solution: If $N$ beats occur each second, the interval between them is $\dfrac{1}{N}$ second.
Q8 — Beats · medium · theory
The phenomenon of beats is a direct consequence of:
A. The principle of superposition of waves ✓ Correct
B. Total internal reflection
C. The Doppler effect
D. The photoelectric effect
Solution: Adding two sinusoids of nearly equal frequency gives a carrier at the mean frequency modulated by a slowly varying envelope.
Q9 — Beats · easy · numerical
Two tuning forks of frequencies $256\text{ Hz}$ and $260\text{ Hz}$ are sounded together. The number of beats heard per second is:
A. $2$
B. $8$
C. $4$ ✓ Correct
D. $258$
Solution: Beat frequency $= |260 - 256| = 4\text{ per second}$.
Q10 — Beats · medium · numerical
Two sound waves of frequencies $340\text{ Hz}$ and $344\text{ Hz}$ propagate in air. The time interval between two successive maxima of intensity is:
A. $0.5\text{ s}$
B. $0.25\text{ s}$ ✓ Correct
C. $0.125\text{ s}$
D. $4\text{ s}$
Solution: Beat frequency $= 4\text{ Hz}$, so the beat period is $\dfrac{1}{4} = 0.25\text{ s}$.
Q11 — Beats · hard · numerical
Two tuning forks give $5$ beats per second. On loading the fork of higher frequency with wax, the beat frequency falls to $2$ per second. If the other fork is $256\text{ Hz}$, the original frequency of the loaded fork was:
A. $258\text{ Hz}$
B. $261\text{ Hz}$ ✓ Correct
C. $251\text{ Hz}$
D. $254\text{ Hz}$
Solution: Waxing lowers the higher frequency towards $256\text{ Hz}$, and the beats do decrease. Hence that fork was above $256$: $256 + 5 = 261\text{ Hz}$.
Q12 — Beats · easy · numerical
Two tuning forks of frequencies $512\text{ Hz}$ and $508\text{ Hz}$ are sounded together. The beat frequency is:
A. $510\text{ Hz}$
B. $1020\text{ Hz}$
C. $4\text{ Hz}$ ✓ Correct
D. $2\text{ Hz}$
Solution: Beat frequency $= |512 - 508| = 4\text{ Hz}$.
Q13 — Beats · easy · numerical
The beat period produced by two sources is $0.2\text{ s}$. The beat frequency is:
A. $10\text{ Hz}$
B. $5\text{ Hz}$ ✓ Correct
C. $0.2\text{ Hz}$
D. $2\text{ Hz}$
Solution: Beat frequency is the reciprocal of the beat period: $\dfrac{1}{0.2} = 5\text{ Hz}$.
Q14 — Beats · medium · numerical
A tuning fork of frequency $300\text{ Hz}$ produces $3$ beats per second with an unknown fork. The frequency of the unknown fork is:
A. $297\text{ Hz}$ or $303\text{ Hz}$ ✓ Correct
B. $300\text{ Hz}$ only
C. $303\text{ Hz}$ only
D. $150\text{ Hz}$ or $600\text{ Hz}$
Solution: The beat frequency gives only the magnitude of the difference, so the unknown fork may lie on either side: $300 \pm 3$.
Q15 — Beats · medium · numerical
Two tuning forks of frequencies $200\text{ Hz}$ and $205\text{ Hz}$ are sounded together. The number of beats heard in $4\text{ seconds}$ is:
A. $10$
B. $20$ ✓ Correct
C. $5$
D. $40$
Solution: Beat frequency $= 5\text{ Hz}$, so in $4\text{ s}$ there are $5 \times 4 = 20$ beats.
Q16 — Beats · easy · numerical
Two sources produce a beat frequency of $6\text{ Hz}$. The number of beats heard in one minute is:
A. $3600$
B. $6$
C. $360$ ✓ Correct
D. $60$
Solution: $6$ beats per second for $60\text{ s}$ gives $6 \times 60 = 360$ beats.
Q17 — Beats · medium · numerical
Two tuning forks of frequencies $256\text{ Hz}$ and $252\text{ Hz}$ are sounded together. The beat period is:
A. $4\text{ s}$
B. $0.25\text{ s}$ ✓ Correct
C. $0.125\text{ s}$
D. $0.5\text{ s}$
Solution: Beat frequency $= 4\text{ Hz}$, so the beat period is $0.25\text{ s}$.
Q18 — Beats · hard · numerical
A fork $A$ of frequency $400\text{ Hz}$ gives $4$ beats per second with a fork $B$. On loading $B$ with wax, the beat frequency rises to $6$ per second. The original frequency of $B$ was:
A. $396\text{ Hz}$ ✓ Correct
B. $406\text{ Hz}$
C. $404\text{ Hz}$
D. $394\text{ Hz}$
Solution: Waxing lowers $B$. If $B$ had been $404\text{ Hz}$ it would move towards $400$ and the beats would fall. Since they rose, $B$ was below $A$, at $396\text{ Hz}$.
Q19 — Beats · easy · numerical
Two sources of frequencies $1000\text{ Hz}$ and $1006\text{ Hz}$ are sounded together. The beat frequency is:
A. $2006\text{ Hz}$
B. $6\text{ Hz}$ ✓ Correct
C. $1003\text{ Hz}$
D. $3\text{ Hz}$
Solution: Beat frequency $= |1006 - 1000| = 6\text{ Hz}$.
Q20 — Beats · hard · numerical
A fork of frequency $288\text{ Hz}$ gives $4$ beats per second with another fork. On filing the second fork, the beat frequency falls to $2$ per second. The original frequency of the second fork was:
A. $292\text{ Hz}$
B. $284\text{ Hz}$ ✓ Correct
C. $290\text{ Hz}$
D. $286\text{ Hz}$
Solution: Filing raises the frequency of the second fork. Since the beats decreased, it must have been moving towards $288\text{ Hz}$ from below, so it was $284\text{ Hz}$.