# Sound Class 8 Numerical

Last updated on November 3rd, 2023 at 11:22 am

In this post, we will solve **Numerical problems based on the Sound chapter of class 8 science**. Here, first, we will go through the required *formulas* & other related equations, and then attempt the numerical questions. Solution & answer is given for all the numerical questions.

**Related Post**: **Light class 8 numerical solved**

**Formulas Used**

Frequency = 1 / Time Period

f = 1/TTime Period = 1/frequency

T = 1/fDistance traveled by a sound wave = speed of sound x time duration

Frequency = number of oscillations per second

Time period = time required for a single oscillation

The frequency is measured in Hertz. If one cycle is completed in one second, the frequency is said to be 1 Hz. For example, say, a tuning fork has a frequency of 512 Hz. It means that the fork vibrates 512 times per second. These vibrations will set 512 cycles of compression and rarefaction in the air, per second. Thus, the sound generated by the fork will have a frequency of 512 Hz.

**Sound Class 8 Numerical Questions and Answers** [solved]

Example 1.An object is vibrating at 50 hertz. What is its time period?

**Solution:**

If an object is vibrating at 50 hertz

50 = 1/T

=> T = 1/50 = 0.02 sec

Its time period will be 0.02 sec.

Example 2.1 hertz is equal to how many vibrations per minute?

**Solution:**

1 Hz= 1 vibration/second

1 minute=60 seconds,

hence 1 hertz = 60 vibrations /minute

Example 3: Lightning can be seen the moment it occurs. Paheli observes lightning in her area. She hears the sound 5 s after she observed lightning. How far is she from the place where lightning occurs? (speed of sound =330 m/s).

**Solution**

Distance=330 m/s x 5s

=1650 m

Example4: A simple pendulum makes 10 oscillations in 20 seconds. What is the time period and frequency of its oscillation?

**Solution**

Time period = time required for one oscillation =20/10 s = 2 s

frequency = 1/Time Period = 1/2 = 0.5 oscillations/sec = 0.5 Hz

**Sound Numerical (Extra questions)**

5 ] If the frequency of a tuning fork is 512 Hz, then how many times it vibrates in 5 seconds?

**Solution**:

Frequency = 512 Hz.

This means the tuning fork vibrates 512 times in 1 second.

So, in 5 seconds, it will vibrate = 5×512 times = **2560 times.**

6 ] If the frequency of a tuning fork is 512 Hz, then how many cycles of compression and rarefaction will happen in the air per second?

**Solution:**

Number of **cycles** of compression and rarefaction in the air per second = frequency

So. in this case, per second 512 **cycles** of compression and rarefaction in the air will happen.

7 ] The total number of compressions and rarefactions produced per second in a sound wave is 1000. Find the frequency of the sound wave.

Solution:

The total number of compressions and rarefactions produced per second in a sound wave = 1000

=> The number of compressions produced per second = The number of rarefactions produced per second = 1000/2 =500

Hence, the number of **cycles** of compression and rarefaction in the air per second =500 [each cycle consists of 1 compression and 1 rarefaction]

We know that, the number of **cycles** of compression and rarefaction in the air per second = frequency

Therefore, the frequency of the sound wave = 500 Hz.

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