Physics Aptitude Test for Class 11 on “Waves – Beats”.
1. Consider the phenomenon of beats & select the correct statement.
a) It occurs when frequencies are close or equal
b) It correspond to the change in intensity we hear due to interference of 2 waves having close frequencies
c) It corresponds to a constant intensity sound heard during wave interference
d) Beat frequency is variable
Answer: b
Clarification: When two waves with close, but not equal, frequencies interfere, beats are produced. They have a frequency equal to the average frequency of the two interfering waves and we recognize them as the variable intensity sound we hear.
2. Two interfering waves have a frequency of 2Hz & 6Hz. What is the beat frequency?
a) 8Hz
b) 4Hz
c) 2Hz
d) 0
Answer: b
Clarification: Beat frequency is equal to the difference between frequencies of interfering waves. In this case, beat frequency
= 6-2 = 4Hz.
3. Let’s say we can hear a beat frequency of 2Hz from two tuning forks. The first tuning fork has a frequency of 100Hz. If wax is applied on the 2nd tuning fork, the beat frequency becomes 4Hz, what is the frequency of the 2nd tuning fork?
a) 100Hz
b) 98Hz
c) 106Hz
d) 104Hz
Answer: b
Clarification: As the initial beat frequency is 2Hz, the 2nd tuning fork can have a frequency of 98Hz or 102Hz. When wax is applied to the second tuning fork its frequency should decrease. On doing so the beat frequency has also increased, so the initial value of the frequency of the 2nd tuning fork must be lower than that of the first tuning fork.
4. Two longitudinal waves: s1 = 3cos(11t) & s2 = 3cos(10t) interfere. What is the amplitude of the resulting wave?
a) 6cos(0.5t)
b) 6sin(0.5t)
c) 6cos(10.5t)
d) 6sin(10.5t)
Answer: a
Clarification: s = s1+s2 = 2(3)*cos(1/2)t*cos(21/2)t
= 6cos(0.5t)cos(10.5t).
The term 6cos(0.5t) corresponds to the amplitude.