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Sound and wave:- Physics numerical problems

 Sound and wave numerical problems for various competition exam.


 Q.1.  If the wavelength of the wave is 5 m and the time period is 2 seconds, then how much time will the wave take to cover a distance of 50 m?  


 A) 25 seconds

 B) 20 seconds

 C) 15 seconds

 D) 10 seconds

 ANSWER= B) 20 seconds

 Explanation:- The time period is called the time taken to complete 1 oscillation.  The wavelength is distance by 1 oscillation.

 Hence, time taken to cover a distance of 5 meters = 2 seconds

 Time taken to cover a distance of 1 meter = 2/5 second

 Time taken to cover a distance of 50 meters = (2/5)×50

 =2×10

 = 20 seconds


 Q.2.  If the frequency is 2 Hz, what will be the wavelength of the wave?  


 A) 1.5 m

 B) 0.5 m

 C) 2.5 m

 D) 3.5 m

 ANSWER= D) 0.5 m

 Explain:-  = 1÷2

 =0.5m

 {Formula :- = 1/n}


 Q.3.  If the frequency is 30 Hz, find the time period.  


 A) 1/30 sec

 B) 0.45 sec

 C) 2/30 sec

 D) 0.57 sec

 ANSWER= A)1/30 sec

 Explain:- Relationship between period T and frequency n :-

 T = 1/n


 Q.4.  Which of the following represents the time period of a simple pendulum:-


 A) T = 2π{√(l÷g)}

 B) T = 4π{√(l÷g)}

 C) T = 3π{√(l÷g)}

 D) T = {√(l÷g)}

 ANSWER= A) T = 2π{√(l÷g)}

 Explain:-


 Q.5.  The time period of a simple pendulum is 2 seconds, then what is the length of the pendulum?  


 A) 1.99 m

 B) 0.99 m

 C) 1.50 m

 D) 2 meters

 ANSWER= B) 0.99 m

 Explain:- T = 2π{√(l÷g)}

 On square off on both sides,

 T² = 4π²(l÷g)

 T²g÷4π²= l

 l = (2²×9.8)÷4π²

 l= 9.8÷π²

 = (9.8×7×7)÷22×22

 =(4.9×7×7)÷11×22

 = 240.1÷242

 =0.99 m


 Q.6.  If the frequency of a vibrating object is 100 per second, what will be its period of oscillation?  


 A) 0.1 sec

 B) 0.2 sec

 C) 0.01 sec

 D) 0.03 sec

 ANSWER= C) 0.01 second

 Explain:- time of oscillation T =(1/n)

 = 1/100

 =0.01  seconds


 Q.7.  What will be the time period of 9.8 m long pendulum?  


 A) 8.21 s

 B) 5.99 s

 C) 7.45 s

 D) 6.2857 s

 ANSWER= D) 6.2857 s

 Explain:- T = 2π(l÷g)

 = 2π(9.8÷9.8)

 =2π

 = 2×3.14

 = 6.28 s


 Q.8.  The length of the simple pendulum at some place is 1m and the period of its oscillation is 1 second, then what will be the value of acceleration due to gravity at that place?  


 A) 28.21 m/s²

 B) 39.51 m/s²

 C) 17.45 m/s²

 D) 31.22 m/s²

 ANSWER= B) 39.51m/s²

 Explain:- From the formula  T = 2π{√(l÷g)},

 g= 4π²l÷T

 g ={ 4(22/7)²×1}÷1

 = (4×22×22)÷49

 = 1936÷49

 = 39.51 m/s²


 Q.9.  What is the length of a second pendulum?  


 A) 1.21 m

 B) 0.9925 m

 C) 0.45 m

 D) 0.2857 m

 ANSWER= B) 0.9925 m

 Explain:- T = 2π{√(l÷g)}

 On square off on both sides,

 T² = 4π²(l÷g)

 T²g÷4π²= l

 Now, the time period of the seconds pendulum is 2  seconds.

 Therefore l = (2²×9.8)÷4π²

 l= 9.8÷π²

 = (9.8×7×7)÷22×22

 =(4.9×7×7)÷11×22

 = 240.1÷242

 =0.99 m


 Q.10.  An object oscillates at simple harmonic motion. If its time period is 8 sec and amplitude is 0.40 m then find the maximum velocity of the object.  


 A) 0.56 m/s

 B) 1.78 m/s

 C) 2.22 m/s

 D) 0.3142 m/s

 ANSWER= D) 0.3142 m/s

 Explain:- time period T= 8s amplitude

 r = 0.40 m

 angular velocity ꙍ

 = 2π÷ T

 = 2π÷8

 = π÷4

 Max Velocity = rω

 = 0.4 × (3.14÷4)

 = 0.4×.785

 = .314 m/s


 Q.11.  What is the time period of the minute hand of a clock? :-


 A) 1 minute

 B) 30 minutes

 C) 120 minutes

 D) 60 minutes

 ANSWER= D) 60 minutes

 Explain:-


 Q.12.  What is the time period of the hour hand:-


 A) 12 hours

 B) 30 minutes

 C) 120 minutes

 D) 60 hours

 ANSWER= A) 12 hours

 Explain:-


 Q.13.  An object is moving at simple harmonic motion.  Its maximum velocity is 1 m/s and maximum acceleration is 1.57 m/s². Calculate the time period:-


 A) 3 seconds

 B) 1 second

 C) 4 seconds

 D) 2 seconds

 ANSWER= C) 4 seconds

 Explain:- Maximum velocity = rω

 Maximum acceleration =ω² r

 ω²r÷ωr =(1.57÷1)

 = 1.57

 Now angular velocity = 2π÷T

 T= 2π÷ω

 = (2×3.14)÷1.57

 = 2×2  = 4 seconds


 Q.14.  The frequency of the wave is 2 Hz and the wavelength is 4 m, then find the velocity of the wave:-


 A) 6.5 m/s

 B) 8 m/s

 C) 6 m/s

 D) 11 m/s

 ANSWER= B) 8 m/s

 Explain:- Wave velocity v = nλ

 = 4×2

 = 8 m/s

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