Question

A spring-block oscillator oscillates with a period of 2.5

seconds. What is its frequency? (Round to the nearest
hundredth)
a 0.2 Hz
b 0.6 Hz
C 0.5 Hz
d 0.4 Hz
e 0.3 Hz

Answer 1

So, the frequency of that spring-block oscillator is 0.40 Hz (D).

Introduction

Hi ! Here, I will help you to explain about this question. Before continuing, let us recall, what is the meaning of frequency. Frequency is the number of vibrations that can be performed per second. In general, the frequency can be formulated by:

`\boxed{\sf{\bold{f = (n)/(t)}}}`

With the following condition :

• f = vibration frequency (Hz)
• n = the number of vibrations that happened
• t = interval of the time (s)

However, the frequency can also be affected by the period of the vibration. The period is the time it takes to pass one vibrating phase. Because of their inversely related meanings, the relationship between frequency and period is expressed in the equation :

`\boxed{\sf{\bold{f = (1)/(T)}}}`

With the following condition :

• f = vibration frequency (Hz)
• T = period of the vibration (s)

Problem Solving

We know that :

• T = period of the vibration = 2.5 s

What was asked :

• f = vibration frequency = ... Hz (Round to nearest hundreth)

Step by step :

`\sf{f = (1)/(T)}`

`\sf{f = (1)/(2.5)}`

`\boxed{\sf{f = 0.40 \: Hz}}`

Conclusion

So, the frequency of that spring-block oscillator is 0.40 Hz (D).