Question

Find an expression for the block's oscillation frequency f in terms of the frequencies f1 and f2 at which it would oscillate if attached to spring 1 or spring 2 alone.

Answer 1

The expression for the oscillation frequency of a block attached to two springs combined in parallel in terms of the individual oscillation frequencies f1 and
`f2 is f = √(f1² + f2²).`

Step-by-step explanation:

The question deals with the oscillation frequency of a block attached to two different springs. When a mass is attached to a single spring, the oscillation frequency (f) of the simple harmonic motion can be calculated using the formula f = 1/T, where T is the period of oscillation.

For a simple harmonic oscillator, the period T is given by
`T = 2π√(m/k)`, where m is the mass attached to the spring, and k is the spring constant. Thus, the frequency is
`f = 1/(2π√(m/k)).`

When two springs (spring 1 and spring 2) are involved, if the block was attached to spring 1 alone, it would oscillate with frequency f1, and if attached to spring 2 alone, it would oscillate with frequency f2. If the springs are combined in parallel, the effective spring constant (k_total) is the sum of the individual spring constants (k1 + k2). Hence, the expression for the block's oscillation frequency f when attached to both springs becomes:

`f = 1/(2π√(m/(k1 + k2)))`

However, to express f directly in terms of f1 and f2, we use the relationship between spring constant and frequency (k = m(2πf)²). Substituting these expressions into the formula for f, we get:

`f = 1/(2π√(m/(m(2πf1)² + m(2πf2)²)))`

`f = 1/(2π√(1/(4π²(f1² + f2²))))`

`f = √(f1² + f2²)`

This is the desired expression for the block's oscillation frequency f in terms of the frequencies f1 and f2.