Any real spring has mass. If this mass is taken into account, explain qualitatively how this will affect the period of oscillation of a spring-block system.

Question

Jitender Pal · Answer

Time period (T) of vibrations varies directly as the square root of the mass (m) of body attached to the spring and inversely as the square root of the force constant of the spring (k).
So, T = 2π √m/k
If the spring is massless then the time period of the oscillation will be,
T = 2π√m/k
When the mass of the spring (m_s) is considered then, the total mass of the system will be (m + m_s),
To obtain the new time period of the oscillation of the system (Tʹ), substitute Tʹ for T and (m + m_s) for m in the equation T = 2π √m/k,
Tʹ = 2π √( m + m_s) /k …… (1)
The value of spring constant will not be changed, because it is always constant.
From equation (1) we observed that, if the mass of the real spring is taken into account, then the period of the oscillation of a spring block system will be increased.

Sherien Gull Sherien Gull · Answer

The spring will vibrate more slowly when it's mass is taken into consideration than when it's mass is neglected.

Muhammad Abdullah · Answer

As we know that the time period of a spring mass system is given by T= 2π√(m/k) If we take mass of spring into account then m'=m+M Here m'=mass of block And M=mass of spring So T'=2π√(m+M)/k So by this equation we can observe that the time period will be increase ,so the spring will vibrate more slowly when it's mass is taken into consideration than when it's mass is neglected So T'>T

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Any real spring has mass. If this mass is taken into account, explain qualitatively how this will affect the period of oscillation of a spring-block system.

Any real spring has mass. If this mass is taken into account, explain qualitatively how this will affect the period of oscillation of a spring-block system.

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