Question

time period of oscillations is T. The correct relation is
a) T^2 = T1^2 + T2^2
B) T^-2 = T1^-2 + T2^-2
C) T^-1 = T1^-1 + T2^-1
D) T = T1 + T2​

Answer 1

Answer:

T^2=T1^-2+T2^-1

B option

Hope it helps

Answer 2

Q]____?

b) T^-2 = T1^-2 + T2^-2

Explanation

Problem solving strategy calculate the effective force constant of parallel spring then by putting the values of time period T = 2π √M /K, we get the new time period of spring.

we can write time period for a vertical spring- block system as

T = 2π √l /g

Here and is extension in the spring when the mass m is suspended from the spring

this can be seen as under :

Kl = mg. ... [in equilibrium position ]

=> m/k = l / g

∴ T = 2π √m/ k

∴ T 1 = 2π √m/k1

=> k1 = 4π^2 [ m /T1^2] ...[i]

T2 = 2π √m/k2

=> k2 = 4π ^2 [m / T2^2 ] ...[ii]

Since, springs are in parallel, effective force constant

K = K1 + K2

T = 2π √( m / k1 +k2 )

=> k1 + k2 = 4π^2 m /T^2 ...[iii]

Substituting values of k1 and k2 from Eqs [i] and [ii], we get

4π^2 (m/ T1^2) + 4π^2 (m/ T2^2) = 4π^2 (m/ T^2)

=> 1/T^2 = 1 / T1^2 + 1 / T2^2

or, T^-2 = T1^-2 + T2^-2

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