GATE-2018 (24) :Wyllie – Time average equation

CSIR-NET 2012 (June) 104 :

(A). If ω and ω0 represent the seismic signal frequencies and resonant frequencies, respectively, of a seismometer, in which of the following case, it would behave as a displacement meter.

A) ω₀ >>ω             B) ω₀ << ω            C) ω₀ =ω       D) ω₀ =1/ω

Answer: B

Explanation:

1. A long period seismometer is an instrument in which the resonant frequency (ω₀) is very low. The lag between the seismometer and the ground motion becomes zero and the amplitude of the seismometer displacement becomes equal to the amplified ground displacement. It is sometimes called a displacement meter.

2. It is usually designed to record seismic signals with frequencies in the range 0.01 Hz to 0.1 Hz which is ‘periods’ of 100 seconds down to 10 seconds.

3. Long-period seismometer also called displacement meter, therefore resonant frequency(ω₀) is very low compared to seismic signal frequency(ω).

Extra information:

Displacement meter:

$u=\frac{A\omega^2}{\omega_0^2 +\omega^2} \cos(\omega t-\triangle)---(1)$ 

Here $ \omega_0^2 << \omega^2$

$u=Acos\omega t=q$

Accelerometer: 

$u=\frac{A\omega^2}{\omega_0^2 +\omega^2} \cos(\omega t-\triangle)---(1)$

Here $ \omega_0^2 >> \omega^2$

$u=\frac{1}{\omega_0^2}Acos\omega t=q$

$\frac{\partial u}{\partial t}=-\frac{1}{\omega_0^2}A \omega sin\omega t$

$\frac{\partial u^2}{\partial^2 t}=-\frac{1}{\omega_0^2}A \omega^2 cos\omega t$

we know that 

$u=Acos\omega t=q$

$\frac{\partial u}{\partial t}=-A \omega sin\omega t=\dot{q}$

$\frac{\partial u^2}{\partial^2 t}=-A \omega^2 cos\omega t =\ddot{q}$

Therefore,

$\frac{\partial u^2}{\partial^2 t}=-\frac{1}{\omega_0^2} \ddot{q}$

Reference: Fundamentals of Geophysics, William Lowrie.

(C). The velocity of tsunami wave traveling through an open ocean of depth 5km is about

A) 220 m/s        B) 20 m/s        C) 2 m/s       D) 800 m/s

Solution:

The velocity of the tsunami wave (V)= √(g*d)

g= acceleration due to gravity = 9.8 m/s

d= depth = 5km

                    = 5000m

V= √ (9.8 * 5000)

  = √(49,000)

  =221.359 m/s

Therefore the velocity of tsunami wave (V) = 220 m/s