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Standard XII

Physics

Center of Mass as an Average Point

A spring bala...

Question

A spring balance has a scale that reads from $0$ to $50$ kg. The length of the scale is $20$ cm. A block of mass m is suspended from this balance, when displaced and released, it oscillates with a period $0.5$ s. The value of m is (Take g = 10 m $s^{- 2})$

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Solution

The correct option is C $16$ kg
The $20$ cm length of the scaler reads upto $50$ kg.
$∴ F = m g = (50 k g) (10 m s^{- 2}) = 500 N$ and $x = 20 c m = 0.2 m$
$∴$ Spring constant, $k = \frac{F}{x} = \frac{500 N}{0.2 m} = 2500 N m^{- 1}$
As $T = 2 π \sqrt{\frac{m}{k}}$
Squaring both sides, we get
$T^{2} = \frac{4 π^{2} m}{k}$
$m = \frac{T^{2} k}{4 π} = \frac{(0.5 s)^{2} \times (2500 N m^{- 1})}{4 \times (3.14)^{2}} = 16 k g$

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A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20cm. A block of mass m is suspended from this balance, when displaced and released, it oscillates with a period 0.5s. The value of m is (Take g = 10 m s−2)

The correct option is C 16kgThe 20cm length of the scaler reads upto 50kg.∴F=mg=(50kg)(10ms−2)=500N and x=20cm=0.2m∴ Spring constant, k=Fx=500N0.2m=2500Nm−1As T=2π√mkSquaring both sides, we getT2=4π2mkm=T2k4π=(0.5s)2×(2500Nm−1)4×(3.14)2=16kg

A spring balance has a scale that reads from $0$ to $50$ kg. The length of the scale is $20$ cm. A block of mass m is suspended from this balance, when displaced and released, it oscillates with a period $0.5$ s. The value of m is (Take g = 10 m $s^{- 2})$