Question

electric vibrator produces ripples in a ripple tank such that the distance between one crest and one trough is 4 cm. If the vibrations are produced at the rate of 4800 vibrations per minute, calculate the time period and velocity of the waves. (0.0125 s; 6.4 m.s-1)

Answer 1

Given

• Distance between crest and trough = 4 cm
• Per minute number of vibrations= 4800

To Find

• Time period
• Velocity of the waves

Solution

☯ Frequency = Number of vibrations per second

• Frequency = 4800/60 = 80 Hz

☯ Wavelength = Distance Between crest and trough × 2

• Wavelength = 4 × 2 = 8 cm

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✭ According to the Question :

→ Velocity = F × λ

• F = Frequency = 80 Hz
• λ = Wavelength = 8 cm = 8/100 = 0.08 m

→ Velocity = 80 × 0.08

→ Velocity = 6.4 m/s

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✭ Time Period :

→ Time = 1/Frequency

→ Time = 1/80

→ Time = 0.0125 sec

Answer 2

Solution

To find

•Velocity of the waves

•Time period

Frequency =Number vibration persecond

•Frequency =4800/69=80Hz

Wavelength =Distance between

cest and through ×2

Wavelength=4×2=8cm

According to the question

Velocity=F×λ

f=Frequency =80Hz

λ=Wavelength=8cm=8/100=0.08

Velocity=80×0.08

Velocity=6.4m/s

Time period

Time =1/Frequency

Time=1/80

Time=0.0125sec